2.2 Linear Equations in One Variable
Learning objectives.
In this section, you will:
- Solve equations in one variable algebraically.
- Solve a rational equation.
- Find a linear equation.
- Given the equations of two lines, determine whether their graphs are parallel or perpendicular.
- Write the equation of a line parallel or perpendicular to a given line.
Caroline is a full-time college student planning a spring break vacation. To earn enough money for the trip, she has taken a part-time job at the local bank that pays $15.00/hr, and she opened a savings account with an initial deposit of $400 on January 15. She arranged for direct deposit of her payroll checks. If spring break begins March 20 and the trip will cost approximately $2,500, how many hours will she have to work to earn enough to pay for her vacation? If she can only work 4 hours per day, how many days per week will she have to work? How many weeks will it take? In this section, we will investigate problems like this and others, which generate graphs like the line in Figure 1 .
Solving Linear Equations in One Variable
A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations.
We begin by classifying linear equations in one variable as one of three types: identity, conditional, or inconsistent. An identity equation is true for all values of the variable. Here is an example of an identity equation.
The solution set consists of all values that make the equation true. For this equation, the solution set is all real numbers because any real number substituted for x x will make the equation true.
A conditional equation is true for only some values of the variable. For example, if we are to solve the equation 5 x + 2 = 3 x − 6 , 5 x + 2 = 3 x − 6 , we have the following:
The solution set consists of one number: { − 4 } . { − 4 } . It is the only solution and, therefore, we have solved a conditional equation.
An inconsistent equation results in a false statement. For example, if we are to solve 5 x − 15 = 5 ( x − 4 ) , 5 x − 15 = 5 ( x − 4 ) , we have the following:
Indeed, −15 ≠ −20. −15 ≠ −20. There is no solution because this is an inconsistent equation.
Solving linear equations in one variable involves the fundamental properties of equality and basic algebraic operations. A brief review of those operations follows.
Linear Equation in One Variable
A linear equation in one variable can be written in the form
where a and b are real numbers, a ≠ 0. a ≠ 0.
Given a linear equation in one variable, use algebra to solve it.
The following steps are used to manipulate an equation and isolate the unknown variable, so that the last line reads x = _________, x = _________, if x is the unknown. There is no set order, as the steps used depend on what is given:
- We may add, subtract, multiply, or divide an equation by a number or an expression as long as we do the same thing to both sides of the equal sign. Note that we cannot divide by zero.
- Apply the distributive property as needed: a ( b + c ) = a b + a c . a ( b + c ) = a b + a c .
- Isolate the variable on one side of the equation.
- When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.
Solving an Equation in One Variable
Solve the following equation: 2 x + 7 = 19. 2 x + 7 = 19.
This equation can be written in the form a x + b = 0 a x + b = 0 by subtracting 19 19 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.
The solution is 6.
Solve the linear equation in one variable: 2 x + 1 = −9. 2 x + 1 = −9.
Solving an Equation Algebraically When the Variable Appears on Both Sides
Solve the following equation: 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) . 4 ( x −3 ) + 12 = 15 −5 ( x + 6 ) .
Apply standard algebraic properties.
This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x = − 5 3 . x = − 5 3 .
Solve the equation in one variable: −2 ( 3 x − 1 ) + x = 14 − x . −2 ( 3 x − 1 ) + x = 14 − x .
Solving a Rational Equation
In this section, we look at rational equations that, after some manipulation, result in a linear equation. If an equation contains at least one rational expression, it is a considered a rational equation .
Recall that a rational number is the ratio of two numbers, such as 2 3 2 3 or 7 2 . 7 2 . A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.
Rational equations have a variable in the denominator in at least one of the terms. Our goal is to perform algebraic operations so that the variables appear in the numerator. In fact, we will eliminate all denominators by multiplying both sides of the equation by the least common denominator (LCD).
Finding the LCD is identifying an expression that contains the highest power of all of the factors in all of the denominators. We do this because when the equation is multiplied by the LCD, the common factors in the LCD and in each denominator will equal one and will cancel out.
Solve the rational equation: 7 2 x − 5 3 x = 22 3 . 7 2 x − 5 3 x = 22 3 .
We have three denominators; 2 x , 3 x , 2 x , 3 x , and 3. The LCD must contain 2 x , 3 x , 2 x , 3 x , and 3. An LCD of 6 x 6 x contains all three denominators. In other words, each denominator can be divided evenly into the LCD. Next, multiply both sides of the equation by the LCD 6 x . 6 x .
A common mistake made when solving rational equations involves finding the LCD when one of the denominators is a binomial—two terms added or subtracted—such as ( x + 1 ) . ( x + 1 ) . Always consider a binomial as an individual factor—the terms cannot be separated. For example, suppose a problem has three terms and the denominators are x , x , x − 1 , x − 1 , and 3 x − 3. 3 x − 3. First, factor all denominators. We then have x , x , ( x − 1 ) , ( x − 1 ) , and 3 ( x − 1 ) 3 ( x − 1 ) as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of ( x − 1 ) . ( x − 1 ) . The x x in the first denominator is separate from the x x in the ( x − 1 ) ( x − 1 ) denominators. An effective way to remember this is to write factored and binomial denominators in parentheses, and consider each parentheses as a separate unit or a separate factor. The LCD in this instance is found by multiplying together the x , x , one factor of ( x − 1 ) , ( x − 1 ) , and the 3. Thus, the LCD is the following:
So, both sides of the equation would be multiplied by 3 x ( x − 1 ) . 3 x ( x − 1 ) . Leave the LCD in factored form, as this makes it easier to see how each denominator in the problem cancels out.
Another example is a problem with two denominators, such as x x and x 2 + 2 x . x 2 + 2 x . Once the second denominator is factored as x 2 + 2 x = x ( x + 2 ) , x 2 + 2 x = x ( x + 2 ) , there is a common factor of x in both denominators and the LCD is x ( x + 2 ) . x ( x + 2 ) .
Sometimes we have a rational equation in the form of a proportion; that is, when one fraction equals another fraction and there are no other terms in the equation.
We can use another method of solving the equation without finding the LCD: cross-multiplication. We multiply terms by crossing over the equal sign.
Multiply a ( d ) a ( d ) and b ( c ) , b ( c ) , which results in a d = b c . a d = b c .
Any solution that makes a denominator in the original expression equal zero must be excluded from the possibilities.
Rational Equations
A rational equation contains at least one rational expression where the variable appears in at least one of the denominators.
Given a rational equation, solve it.
- Factor all denominators in the equation.
- Find and exclude values that set each denominator equal to zero.
- Find the LCD.
- Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
- Solve the remaining equation.
- Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.
Solving a Rational Equation without Factoring
Solve the following rational equation:
We have three denominators: x , x , 2 , 2 , and 2 x . 2 x . No factoring is required. The product of the first two denominators is equal to the third denominator, so, the LCD is 2 x . 2 x . Only one value is excluded from a solution set, 0. Next, multiply the whole equation (both sides of the equal sign) by 2 x . 2 x .
The proposed solution is −1, which is not an excluded value, so the solution set contains one number, −1 , −1 , or { −1 } { −1 } written in set notation.
Solve the rational equation: 2 3 x = 1 4 − 1 6 x . 2 3 x = 1 4 − 1 6 x .
Solving a Rational Equation by Factoring the Denominator
Solve the following rational equation: 1 x = 1 10 − 3 4 x . 1 x = 1 10 − 3 4 x .
First find the common denominator. The three denominators in factored form are x , 10 = 2 ⋅ 5 , x , 10 = 2 ⋅ 5 , and 4 x = 2 ⋅ 2 ⋅ x . 4 x = 2 ⋅ 2 ⋅ x . The smallest expression that is divisible by each one of the denominators is 20 x . 20 x . Only x = 0 x = 0 is an excluded value. Multiply the whole equation by 20 x . 20 x .
The solution is 35 2 . 35 2 .
Solve the rational equation: − 5 2 x + 3 4 x = − 7 4 . − 5 2 x + 3 4 x = − 7 4 .
Solving Rational Equations with a Binomial in the Denominator
Solve the following rational equations and state the excluded values:
- ⓐ 3 x − 6 = 5 x 3 x − 6 = 5 x
- ⓑ x x − 3 = 5 x − 3 − 1 2 x x − 3 = 5 x − 3 − 1 2
- ⓒ x x − 2 = 5 x − 2 − 1 2 x x − 2 = 5 x − 2 − 1 2
The denominators x x and x − 6 x − 6 have nothing in common. Therefore, the LCD is the product x ( x − 6 ) . x ( x − 6 ) . However, for this problem, we can cross-multiply.
The solution is 15. The excluded values are 6 6 and 0. 0.
The LCD is 2 ( x − 3 ) . 2 ( x − 3 ) . Multiply both sides of the equation by 2 ( x − 3 ) . 2 ( x − 3 ) .
The solution is 13 3 . 13 3 . The excluded value is 3. 3.
The least common denominator is 2 ( x − 2 ) . 2 ( x − 2 ) . Multiply both sides of the equation by x ( x − 2 ) . x ( x − 2 ) .
The solution is 4. The excluded value is 2. 2.
Solve − 3 2 x + 1 = 4 3 x + 1 . − 3 2 x + 1 = 4 3 x + 1 . State the excluded values.
Solving a Rational Equation with Factored Denominators and Stating Excluded Values
Solve the rational equation after factoring the denominators: 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . 2 x + 1 − 1 x − 1 = 2 x x 2 − 1 . State the excluded values.
We must factor the denominator x 2 −1. x 2 −1. We recognize this as the difference of squares, and factor it as ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Thus, the LCD that contains each denominator is ( x − 1 ) ( x + 1 ) . ( x − 1 ) ( x + 1 ) . Multiply the whole equation by the LCD, cancel out the denominators, and solve the remaining equation.
The solution is −3. −3. The excluded values are 1 1 and −1. −1.
Solve the rational equation: 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 . 2 x − 2 + 1 x + 1 = 1 x 2 − x − 2 .
Finding a Linear Equation
Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y = m x + b , y = m x + b , where m = slope m = slope and b = y -intercept . b = y -intercept . Let us begin with the slope.
The Slope of a Line
The slope of a line refers to the ratio of the vertical change in y over the horizontal change in x between any two points on a line. It indicates the direction in which a line slants as well as its steepness. Slope is sometimes described as rise over run.
If the slope is positive, the line slants to the right. If the slope is negative, the line slants to the left. As the slope increases, the line becomes steeper. Some examples are shown in Figure 2 . The lines indicate the following slopes: m = −3 , m = −3 , m = 2 , m = 2 , and m = 1 3 . m = 1 3 .
The slope of a line, m , represents the change in y over the change in x. Given two points, ( x 1 , y 1 ) ( x 1 , y 1 ) and ( x 2 , y 2 ) , ( x 2 , y 2 ) , the following formula determines the slope of a line containing these points:
Finding the Slope of a Line Given Two Points
Find the slope of a line that passes through the points ( 2 , −1 ) ( 2 , −1 ) and ( −5 , 3 ) . ( −5 , 3 ) .
We substitute the y- values and the x- values into the formula.
The slope is − 4 7 . − 4 7 .
It does not matter which point is called ( x 1 , y 1 ) ( x 1 , y 1 ) or ( x 2 , y 2 ) . ( x 2 , y 2 ) . As long as we are consistent with the order of the y terms and the order of the x terms in the numerator and denominator, the calculation will yield the same result.
Find the slope of the line that passes through the points ( −2 , 6 ) ( −2 , 6 ) and ( 1 , 4 ) . ( 1 , 4 ) .
Identifying the Slope and y- intercept of a Line Given an Equation
Identify the slope and y- intercept, given the equation y = − 3 4 x − 4. y = − 3 4 x − 4.
As the line is in y = m x + b y = m x + b form, the given line has a slope of m = − 3 4 . m = − 3 4 . The y- intercept is b = −4. b = −4.
The y -intercept is the point at which the line crosses the y- axis. On the y- axis, x = 0. x = 0. We can always identify the y- intercept when the line is in slope-intercept form, as it will always equal b. Or, just substitute x = 0 x = 0 and solve for y.
The Point-Slope Formula
Given the slope and one point on a line, we can find the equation of the line using the point-slope formula.
This is an important formula, as it will be used in other areas of college algebra and often in calculus to find the equation of a tangent line. We need only one point and the slope of the line to use the formula. After substituting the slope and the coordinates of one point into the formula, we simplify it and write it in slope-intercept form.
Given one point and the slope, the point-slope formula will lead to the equation of a line:
Finding the Equation of a Line Given the Slope and One Point
Write the equation of the line with slope m = −3 m = −3 and passing through the point ( 4 , 8 ) . ( 4 , 8 ) . Write the final equation in slope-intercept form.
Using the point-slope formula, substitute −3 −3 for m and the point ( 4 , 8 ) ( 4 , 8 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .
Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.
Given m = 4 , m = 4 , find the equation of the line in slope-intercept form passing through the point ( 2 , 5 ) . ( 2 , 5 ) .
Finding the Equation of a Line Passing Through Two Given Points
Find the equation of the line passing through the points ( 3 , 4 ) ( 3 , 4 ) and ( 0 , −3 ) . ( 0 , −3 ) . Write the final equation in slope-intercept form.
First, we calculate the slope using the slope formula and two points.
Next, we use the point-slope formula with the slope of 7 3 , 7 3 , and either point. Let’s pick the point ( 3 , 4 ) ( 3 , 4 ) for ( x 1 , y 1 ) . ( x 1 , y 1 ) .
In slope-intercept form, the equation is written as y = 7 3 x − 3. y = 7 3 x − 3.
To prove that either point can be used, let us use the second point ( 0 , −3 ) ( 0 , −3 ) and see if we get the same equation.
We see that the same line will be obtained using either point. This makes sense because we used both points to calculate the slope.
Standard Form of a Line
Another way that we can represent the equation of a line is in standard form . Standard form is given as
where A , A , B , B , and C C are integers. The x- and y- terms are on one side of the equal sign and the constant term is on the other side.
Finding the Equation of a Line and Writing It in Standard Form
Find the equation of the line with m = −6 m = −6 and passing through the point ( 1 4 , −2 ) . ( 1 4 , −2 ) . Write the equation in standard form.
We begin using the point-slope formula.
From here, we multiply through by 2, as no fractions are permitted in standard form, and then move both variables to the left aside of the equal sign and move the constants to the right.
This equation is now written in standard form.
Find the equation of the line in standard form with slope m = − 1 3 m = − 1 3 and passing through the point ( 1 , 1 3 ) . ( 1 , 1 3 ) .
Vertical and Horizontal Lines
The equations of vertical and horizontal lines do not require any of the preceding formulas, although we can use the formulas to prove that the equations are correct. The equation of a vertical line is given as
where c is a constant. The slope of a vertical line is undefined, and regardless of the y- value of any point on the line, the x- coordinate of the point will be c .
Suppose that we want to find the equation of a line containing the following points: ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , ( −3 , −5 ) , ( −3 , 1 ) , ( −3 , 3 ) , and ( −3 , 5 ) . ( −3 , 5 ) . First, we will find the slope.
Zero in the denominator means that the slope is undefined and, therefore, we cannot use the point-slope formula. However, we can plot the points. Notice that all of the x- coordinates are the same and we find a vertical line through x = −3. x = −3. See Figure 3 .
The equation of a horizontal line is given as
where c is a constant. The slope of a horizontal line is zero, and for any x- value of a point on the line, the y- coordinate will be c .
Suppose we want to find the equation of a line that contains the following set of points: ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , ( −2 , −2 ) , ( 0 , −2 ) , ( 3 , −2 ) , and ( 5 , −2 ) . ( 5 , −2 ) . We can use the point-slope formula. First, we find the slope using any two points on the line.
Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y -intercept.
The graph is a horizontal line through y = −2. y = −2. Notice that all of the y- coordinates are the same. See Figure 3 .
Finding the Equation of a Line Passing Through the Given Points
Find the equation of the line passing through the given points: ( 1 , −3 ) ( 1 , −3 ) and ( 1 , 4 ) . ( 1 , 4 ) .
The x- coordinate of both points is 1. Therefore, we have a vertical line, x = 1. x = 1.
Find the equation of the line passing through ( −5 , 2 ) ( −5 , 2 ) and ( 2 , 2 ) . ( 2 , 2 ) .
Determining Whether Graphs of Lines are Parallel or Perpendicular
Parallel lines have the same slope and different y- intercepts. Lines that are parallel to each other will never intersect. For example, Figure 4 shows the graphs of various lines with the same slope, m = 2. m = 2.
All of the lines shown in the graph are parallel because they have the same slope and different y- intercepts.
Lines that are perpendicular intersect to form a 90° 90° -angle. The slope of one line is the negative reciprocal of the other. We can show that two lines are perpendicular if the product of the two slopes is −1 : m 1 ⋅ m 2 = −1. −1 : m 1 ⋅ m 2 = −1. For example, Figure 5 shows the graph of two perpendicular lines. One line has a slope of 3; the other line has a slope of − 1 3 . − 1 3 .
Graphing Two Equations, and Determining Whether the Lines are Parallel, Perpendicular, or Neither
Graph the equations of the given lines, and state whether they are parallel, perpendicular, or neither: 3 y = − 4 x + 3 3 y = − 4 x + 3 and 3 x − 4 y = 8. 3 x − 4 y = 8.
The first thing we want to do is rewrite the equations so that both equations are in slope-intercept form.
First equation:
Second equation:
See the graph of both lines in Figure 6
From the graph, we can see that the lines appear perpendicular, but we must compare the slopes.
The slopes are negative reciprocals of each other, confirming that the lines are perpendicular.
Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2 y − x = 10 2 y − x = 10 and 2 y = x + 4. 2 y = x + 4.
Writing the Equations of Lines Parallel or Perpendicular to a Given Line
As we have learned, determining whether two lines are parallel or perpendicular is a matter of finding the slopes. To write the equation of a line parallel or perpendicular to another line, we follow the same principles as we do for finding the equation of any line. After finding the slope, use the point-slope formula to write the equation of the new line.
Given an equation for a line, write the equation of a line parallel or perpendicular to it.
- Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
- Use the slope and the given point with the point-slope formula.
- Simplify the line to slope-intercept form and compare the equation to the given line.
Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point
Write the equation of line parallel to a 5 x + 3 y = 1 5 x + 3 y = 1 and passing through the point ( 3 , 5 ) . ( 3 , 5 ) .
First, we will write the equation in slope-intercept form to find the slope.
The slope is m = − 5 3 . m = − 5 3 . The y- intercept is 1 3 , 1 3 , but that really does not enter into our problem, as the only thing we need for two lines to be parallel is the same slope. The one exception is that if the y- intercepts are the same, then the two lines are the same line. The next step is to use this slope and the given point with the point-slope formula.
The equation of the line is y = − 5 3 x + 10. y = − 5 3 x + 10. See Figure 7 .
Find the equation of the line parallel to 5 x = 7 + y 5 x = 7 + y and passing through the point ( −1 , −2 ) . ( −1 , −2 ) .
Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point
Find the equation of the line perpendicular to 5 x − 3 y + 4 = 0 5 x − 3 y + 4 = 0 and passing through the point ( − 4 , 1 ) . ( − 4 , 1 ) .
The first step is to write the equation in slope-intercept form.
We see that the slope is m = 5 3 . m = 5 3 . This means that the slope of the line perpendicular to the given line is the negative reciprocal, or − 3 5 . − 3 5 . Next, we use the point-slope formula with this new slope and the given point.
Access these online resources for additional instruction and practice with linear equations.
- Solving rational equations
- Equation of a line given two points
- Finding the equation of a line perpendicular to another line through a given point
- Finding the equation of a line parallel to another line through a given point
2.2 Section Exercises
What does it mean when we say that two lines are parallel?
What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?
How do we recognize when an equation, for example y = 4 x + 3 , y = 4 x + 3 , will be a straight line (linear) when graphed?
What does it mean when we say that a linear equation is inconsistent?
When solving the following equation:
2 x − 5 = 4 x + 1 2 x − 5 = 4 x + 1
explain why we must exclude x = 5 x = 5 and x = −1 x = −1 as possible solutions from the solution set.
For the following exercises, solve the equation for x . x .
7 x + 2 = 3 x − 9 7 x + 2 = 3 x − 9
4 x − 3 = 5 4 x − 3 = 5
3 ( x + 2 ) − 12 = 5 ( x + 1 ) 3 ( x + 2 ) − 12 = 5 ( x + 1 )
12 − 5 ( x + 3 ) = 2 x − 5 12 − 5 ( x + 3 ) = 2 x − 5
1 2 − 1 3 x = 4 3 1 2 − 1 3 x = 4 3
x 3 − 3 4 = 2 x + 3 12 x 3 − 3 4 = 2 x + 3 12
2 3 x + 1 2 = 31 6 2 3 x + 1 2 = 31 6
3 ( 2 x − 1 ) + x = 5 x + 3 3 ( 2 x − 1 ) + x = 5 x + 3
2 x 3 − 3 4 = x 6 + 21 4 2 x 3 − 3 4 = x 6 + 21 4
x + 2 4 − x − 1 3 = 2 x + 2 4 − x − 1 3 = 2
For the following exercises, solve each rational equation for x . x . State all x -values that are excluded from the solution set.
3 x − 1 3 = 1 6 3 x − 1 3 = 1 6
2 − 3 x + 4 = x + 2 x + 4 2 − 3 x + 4 = x + 2 x + 4
3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 ) 3 x − 2 = 1 x − 1 + 7 ( x − 1 ) ( x − 2 )
3 x x − 1 + 2 = 3 x − 1 3 x x − 1 + 2 = 3 x − 1
5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3 5 x + 1 + 1 x − 3 = − 6 x 2 − 2 x − 3
1 x = 1 5 + 3 2 x 1 x = 1 5 + 3 2 x
For the following exercises, find the equation of the line using the point-slope formula. Write all the final equations using the slope-intercept form.
( 0 , 3 ) ( 0 , 3 ) with a slope of 2 3 2 3
( 1 , 2 ) ( 1 , 2 ) with a slope of − 4 5 − 4 5
x -intercept is 1, and ( −2 , 6 ) ( −2 , 6 )
y -intercept is 2, and ( 4 , −1 ) ( 4 , −1 )
( −3 , 10 ) ( −3 , 10 ) and ( 5 , −6 ) ( 5 , −6 )
( 1 , 3 ) and ( 5 , 5 ) ( 1 , 3 ) and ( 5 , 5 )
parallel to y = 2 x + 5 y = 2 x + 5 and passes through the point ( 4 , 3 ) ( 4 , 3 )
perpendicular to 3 y = x − 4 3 y = x − 4 and passes through the point ( −2 , 1 ) ( −2 , 1 ) .
For the following exercises, find the equation of the line using the given information.
( − 2 , 0 ) ( − 2 , 0 ) and ( −2 , 5 ) ( −2 , 5 )
( 1 , 7 ) ( 1 , 7 ) and ( 3 , 7 ) ( 3 , 7 )
The slope is undefined and it passes through the point ( 2 , 3 ) . ( 2 , 3 ) .
The slope equals zero and it passes through the point ( 1 , −4 ) . ( 1 , −4 ) .
The slope is 3 4 3 4 and it passes through the point ( 1 , 4 ) ( 1 , 4 ) .
( –1 , 3 ) ( –1 , 3 ) and ( 4 , –5 ) ( 4 , –5 )
For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.
y = 2 x + 7 y = − 1 2 x − 4 y = 2 x + 7 y = − 1 2 x − 4
3 x − 2 y = 5 6 y − 9 x = 6 3 x − 2 y = 5 6 y − 9 x = 6
y = 3 x + 1 4 y = 3 x + 2 y = 3 x + 1 4 y = 3 x + 2
x = 4 y = −3 x = 4 y = −3
For the following exercises, find the slope of the line that passes through the given points.
( 5 , 4 ) ( 5 , 4 ) and ( 7 , 9 ) ( 7 , 9 )
( −3 , 2 ) ( −3 , 2 ) and ( 4 , −7 ) ( 4 , −7 )
( −5 , 4 ) ( −5 , 4 ) and ( 2 , 4 ) ( 2 , 4 )
( −1 , −2 ) ( −1 , −2 ) and ( 3 , 4 ) ( 3 , 4 )
( 3 , −2 ) ( 3 , −2 ) and ( 3 , −2 ) ( 3 , −2 )
For the following exercises, find the slope of the lines that pass through each pair of points and determine whether the lines are parallel or perpendicular.
( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 ) ( −1 , 3 ) and ( 5 , 1 ) ( −2 , 3 ) and ( 0 , 9 )
( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 ) ( 2 , 5 ) and ( 5 , 9 ) ( −1 , −1 ) and ( 2 , 3 )
For the following exercises, express the equations in slope intercept form (rounding each number to the thousandths place). Enter this into a graphing calculator as Y1, then adjust the ymin and ymax values for your window to include where the y -intercept occurs. State your ymin and ymax values.
0.537 x − 2.19 y = 100 0.537 x − 2.19 y = 100
4,500 x − 200 y = 9,528 4,500 x − 200 y = 9,528
200 − 30 y x = 70 200 − 30 y x = 70
Starting with the point-slope formula y − y 1 = m ( x − x 1 ) , y − y 1 = m ( x − x 1 ) , solve this expression for x x in terms of x 1 , y , y 1 , x 1 , y , y 1 , and m m .
Starting with the standard form of an equation A x + B y = C A x + B y = C solve this expression for y y in terms of A , B , C A , B , C and x x . Then put the expression in slope-intercept form.
Use the above derived formula to put the following standard equation in slope intercept form: 7 x − 5 y = 25. 7 x − 5 y = 25.
Given that the following coordinates are the vertices of a rectangle, prove that this truly is a rectangle by showing the slopes of the sides that meet are perpendicular.
( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) ( – 1 , 1 ) , ( 2 , 0 ) , ( 3 , 3 ) and ( 0 , 4 ) ( 0 , 4 )
Find the slopes of the diagonals in the previous exercise. Are they perpendicular?
Real-World Applications
The slope for a wheelchair ramp for a home has to be 1 12 . 1 12 . If the vertical distance from the ground to the door bottom is 2.5 ft, find the distance the ramp has to extend from the home in order to comply with the needed slope.
If the profit equation for a small business selling x x number of item one and y y number of item two is p = 3 x + 4 y , p = 3 x + 4 y , find the y y value when p = $ 453 and x = 75. p = $ 453 and x = 75.
For the following exercises, use this scenario: The cost of renting a car is $45/wk plus $0.25/mi traveled during that week. An equation to represent the cost would be y = 45 + .25 x , y = 45 + .25 x , where x x is the number of miles traveled.
What is your cost if you travel 50 mi?
If your cost were $ 63.75 , $ 63.75 , how many miles were you charged for traveling?
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- Section URL: https://openstax.org/books/college-algebra-2e/pages/2-2-linear-equations-in-one-variable
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| → → Linear equations Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. You can customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. The worksheets suit and courses (grades 6-9). You can choose from SEVEN basic types of equations, ranging from simple to complex, explained below (such as one-step equations, variable on both sides, or having to use the distributive property). Customize the worksheets using the below. Each worksheet is randomly generated and thus unique. The and is placed on the second page of the file. You can generate the worksheets — both are easy to print. To get the PDF worksheet, simply push the button titled " " or " ". To get the worksheet in html format, push the button " " or " ". This has the advantage that you can save the worksheet directly from your browser (choose File → Save) and then in Word or other word processing program. Sometimes the generated worksheet is not exactly what you want. Just try again! To get a different worksheet using the same options:
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Unit 3 – Linear Functions, Equations, and Their AlgebraDirect Variation LESSON/HOMEWORK LESSON VIDEO EDITABLE LESSON EDITABLE KEY Average Rate of Change Forms of a Line Linear Modeling Inverses of Linear Functions Piecewise Linear Functions Systems of Linear Equations (Primarily 3 by 3) Unit Review Unit 3 Review – Linear Functions UNIT REVIEW EDITABLE REVIEW Unit 3 Assessment Form A EDITABLE ASSESSMENT Unit 3 Assessment Form B Unit 3 Assessment Form C Unit 3 Assessment Form D Unit 3 Exit Tickets Unit 3 Mid-Unit Quiz (Through Lesson #4) – Form A Unit 3 Mid-Unit Quiz (Through Lesson #4) – Form B Unit 3 Mid-Unit Quiz (Through Lesson #4) – Form C Unit 3 Mid-Unit Quiz (Through Lesson #4) – Form D U03.AO.01 – Forms of a Line – Desmos Activity EDITABLE RESOURCE U03.AO.02 – Forms of a Line – Teacher Directions U03.AO.03 – Piecewise Linear Function Practice U03.AO.04 – Inverses of Linear Functions – Practice U03.AO.05 – Practice with Linear Modeling Thank you for using eMATHinstruction materials. In order to continue to provide high quality mathematics resources to you and your students we respectfully request that you do not post this or any of our files on any website. Doing so is a violation of copyright. Using these materials implies you agree to our terms and conditions and single user license agreement . The content you are trying to access requires a membership . If you already have a plan, please login. If you need to purchase a membership we offer yearly memberships for tutors and teachers and special bulk discounts for schools. Sorry, the content you are trying to access requires verification that you are a mathematics teacher. Please click the link below to submit your verification request. 
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selected template will load here This action is not available.  Chapter 2: Linear Equations
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \) \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\) \( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\) \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\) \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vectorC}[1]{\textbf{#1}} \) \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \) \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \) \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
 Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. Chapter 2: Linear Equations 2.2 Solving Linear EquationsWhen working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable. Example 2.2.1 Solve [latex]4x+16=-4[/latex] for [latex]x.[/latex] [latex]\begin{array}{rrrrrl} 4x& +& 16 &=&-4& \\ &&-16&& -16&\text{Subtract 16 from each side} \\ \hline &&\dfrac{4x}{4}& =& \dfrac{-20}{4}&\text{Divide each side by 4}\\ \\ &&x& =& -5 & \text{Solution} \end{array}[/latex] In solving the above equation, notice that the general pattern followed was to do the opposite of the equation. [latex]4x[/latex] was added to 16, so 16 was then subtracted from both sides. The variable [latex]x[/latex] was multiplied by 4, so both sides were divided by 4. Example 2.2.2
For questions 1 to 20, solve each linear equation.
Answer Key 2.2 Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted. Share This Book Game Central
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selected template will load here This action is not available.  1.1: Linear Equations
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \) \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) ( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\) \( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\) \( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\) \( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vectorC}[1]{\textbf{#1}} \) \( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \) \( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \) \( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \) Definition: Linear EquationA linear equation is an equation where the highest exponent on the given variables is one. A linear equation in one variable is an equation with one variable with exponent one, e.g., \[ax+b=c,\nonumber\] where \(a\) is called the coefficient of \(x\), and \(b\) and \(c\) are constant coefficients . Solving linear equations is an important and fundamental skill in algebra. In algebra, we are often presented with a problem where the answer is unknown. This is usually represented by a variable. There will be times when there are multiple unknowns and we use algebra techniques to solve for the variable. Be sure to understand the difference between an expression and an equation. Recall . We simplify expressions and solve equations. Hence, expressions do not contain an equal sign, \(=\), and we only distribute and combine like terms. Equations contain an equal sign, \(=\), and we solve for the variable in addition to distributing and combining like terms. The result to an expression is an expression and the result to an equation is a number (with the exception of when the result is infinitely many solutions or no solution). Verifying SolutionsExample \(\pageindex{1}\). Is \(x = −5\) a solution to the equation \(4x + 16 = −4\)? We substitute \(x = −5\) into the equation and determine if the left side is equal to the right side. \[\begin{array}{rl}4(-5)+16\stackrel{?}{=}-4&\text{Multiply }4(-5) \\ -20+16\stackrel{?}{=}-4&\text{Add }-20+16 \\ -4=-4&\checkmark\text{ True}\end{array}\nonumber\] Hence, \(x = −5\) is a solution to the equation \(4x + 16 = −4\). Example \(\PageIndex{2}\)Is \(x = 3\) a solution to the equation \(4x + 16 = −4\)? We substitute \(x = 3\) into the equation and determine if the left side is equal to the right side. \[\begin{array}{rl}4(3)+16\stackrel{?}{=}-4&\text{Multiply }4(3) \\ 12+16\stackrel{?}{=}-4&\text{Add }12+16 \\ 28\neq -4&X\text{ False}\end{array}\nonumber\] Hence, \(x = 3\) is not a solution to the equation \(4x + 16 = −4\). Example \(\PageIndex{2}\) reiterates that the solution to each equation is unique. Once we see that at one solution makes the equation true, then we look no further. The exception is when we have an identity, which we discuss later in this chapter. One-Step EquationsDepending on the complexity of the problem, this “guess and check” method is not very efficient. Thus, we take a more algebraic approach for solving equations. Here we will focus on equations that only require one step to solve. While these equations often seem very fundamental, it is important to master the pattern for solving these problems so we can solve more complex problems. Addition Property of EquationsGiven an equation, \(a = b\), the following is an equivalent statement: \[a+c=b+c\nonumber\] i.e., we can add any number to one side of the equation as long as we add the same number to the other side. Often, we use this property to isolate the variable. Example \(\PageIndex{3}\)Solve for \(x\): \(x + 7 = −5\) \[\begin{array}{rl}x+7=-5&\text{Isolate }y\text{ by adding the opposite of }7 \\ x+7+\color{blue}{(-7)}\color{black}{}=-5+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ x=-12&\text{Solution}\end{array}\nonumber\] Recall, it is encouraged for the student to check the obtained answer by verifying the solution: \[\begin{array}{rl}\color{blue}{(-12)}\color{black}{}+7\stackrel{?}{=}-5 \\ -5=-5&\checkmark\text{ True}\end{array}\nonumber\] Example \(\PageIndex{4}\)Solve for \(y\): \(4 + y = 8\) \[\begin{array}{rl}4+y=8&\text{Isolate }y\text{ by adding the opposite of }4 \\ 4+y+\color{blue}{(-4)}\color{black}{}=8+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ y=4&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{5}\)Solve for \(y\): \(7 = y + 9\) \[\begin{array}{rl}7=y+9&\text{Isolate }y\text{ by adding the opposite of }9 \\ 7+\color{blue}{(-9)}\color{black}{}=y+9+\color{blue}{(-9)}\color{black}{}&\text{Simplify} \\ -2=y&\text{Rewrite with }y\text{ on the left side} \\ y=-2&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{6}\)Solve for \(x\): \(5 = 8 + x\) \[\begin{array}{rl}5=8+x&\text{Isolate }x\text{ by adding the opposite of }8 \\ 5+\color{blue}{(-8)}\color{black}{}=8+x+\color{blue}{(-8)}\color{black}{}&\text{Simplify} \\ -3=x&\text{Rewrite with }x\text{ on the left side} \\ x=-3&\text{Solution}\end{array}\nonumber\] Be sure to check your answer by verifying the solution! It only take a few seconds and will save you time and future common errors in the long run. Example \(\PageIndex{7}\)Solve for \(y\): \(y − 5 = 4\) \[\begin{array}{rl}y-5=4&\text{Isolate }y\text{ by adding the opposite of }-5 \\ y-5+\color{blue}{(5)}\color{black}{}=4+\color{blue}{(5)}\color{black}{}&\text{Simplify} \\ y=9&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{8}\)Solve for \(y\): \(−10 = y − 7\) \[\begin{array}{rl}-10=y-7&\text{Isolate }y\text{ by adding the opposite of }-7 \\ -10+\color{blue}{(7)}\color{black}{}=y-7+\color{blue}{(7)}\color{black}{}&\text{Simplify} \\ y=-3&\text{Solution}\end{array}\nonumber\] Multiplication Property of EquationsGiven an equation, \(a = b\), the following is an equivalent statement: \[ac=bc,\nonumber\] where \(c\neq 0\), i.e., we can multiply both sides of the equation by a nonzero number. Often, we use this property to isolate the variable when there is a coefficient in front of the variable. Example \(\PageIndex{9}\)Solve for \(y\): \(4y = 20\) \[\begin{array}{rl}4y=20&\text{Isolate }y\text{ by multiplying by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4y=20\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\] In Example \(\PageIndex{9}\) , we multiplied by the reciprocal . The product of a number and its reciprocal is one, i.e., if \(c\) is a number, then its reciprocal is \(\frac{1}{c}\) because \[c\cdot\frac{1}{c}=1\nonumber\] Example \(\PageIndex{10}\)Solve for \(y\): \(−5y = 30\) \[\begin{array}{rl}-5y=30&\text{Isolate }y\text{ by multiplying by the reciprocal of }-5 \\ \color{blue}{-\frac{1}{5}}\color{black}{}\cdot -5y=30\cdot\color{blue}{-\frac{1}{5}}\color{black}{}&\text{Simplify} \\ y=-6&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{11}\)Solve for \(x\): \(−42 = −7x\) \[\begin{array}{rl}-42=-7x&\text{Isolate }x\text{ by multiplying by the reciprocal of }-7 \\ \color{blue}{-\frac{1}{7}}\color{black}{}\cdot -42=-7x\cdot\color{blue}{-\frac{1}{7}}\color{black}{}&\text{Simplify} \\ 6=x&\text{Rewrite with }x\text{ on the left side} \\ x=6&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{12}\)Solve for \(x\): \(\frac{x}{-7}=-2\) \[\begin{array}{rl}\frac{x}{-7}=-2&\text{Isolate }x\text{ by multiplying by the reciprocal of }-\frac{1}{7} \\ \color{blue}{-7}\color{black}{}\cdot\frac{x}{-7}=-2\cdot\color{blue}{-7}\color{black}{}&\text{Simplify} \\ x=14&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{13}\)Solve for \(x\): \(\frac{x}{5}=-3\) \[\begin{array}{rl}\frac{x}{5}=-3&\text{Isolate }x\text{ by multiplying by the reciprocal of }\frac{1}{5} \\ \color{blue}{5}\color{black}{}\cdot\frac{x}{5}=-3\cdot\color{blue}{5}\color{black}{}&\text{Simplify} \\ x=-15&\text{Solution}\end{array}\nonumber\] The processes described above is fundamental for solving equations. Once these processes are mastered, we are given problems that involve multiple steps. These problems may seem more complex, but the processes and patterns are the same. The study of algebra was originally called “ Cossic Art ” from the Latin, meaning the study of “things,” which we now call variables. Two-Step EquationsAfter mastering the techniques for solving one-step equations, we are ready to consider two-step equations. As we solve two-step equations, the important thing to remember is that everything works in reverse. When working with one-step equations, we learned that in order to clear a “plus five” in the equation, we would subtract five (or add its opposite) from both sides. We learned that to clear “divided by seven,” we multiply by seven (or multiply by its reciprocal) on both sides. When solving for our variable, we use order of operations in reverse. This means we will apply the addition property first, then the multiplication property second (then exponents, and, finally, any parenthesis or grouping symbols, but that’s another lesson). Example \(\PageIndex{14}\)Solve for \(y\): \(4y − 20 = −8\) We have one extra term on the same side as the variable term, \(4y\). We will first isolate the variable term, then isolate the variable by applying the properties of equations: \[\begin{array}{rl}4y-20=-8&\text{Add the opposite of }-20\text{ to each side} \\ 4y-20+\color{blue}{20}\color{black}{}=-8+\color{blue}{20}\color{black}{}&\text{Simplify} \\ 4y=12&\text{Multiply by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4y=12\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ y=3&\text{Solution}\end{array}\nonumber\] Let's verify the solution: \[\begin{array}{rl}4\color{blue}{(3)}\color{black}{}-20\stackrel{?}{=}-8 \\ 12-20\stackrel{?}{=}-8 \\ -8=-8&\checkmark\text{ True}\end{array}\nonumber\] Thus, \(y = 3\) is the solution to the equation. The same process is used to solve any two-step equations. Add or subtract first, then multiply or divide. Recall, the method never changes, just problems do. Example \(\PageIndex{15}\)Solve for \(w\): \(5w + 7 = 7\) \[\begin{array}{rl}5w+7=7&\text{Add the opposite of }7\text{ to each side} \\ 5w+7+\color{blue}{(-7)}\color{black}{}=7+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ 5w=0&\text{Multiply by the reciprocal of }5 \\ \color{blue}{\frac{1}{5}}\color{black}{}\cdot 5w=0\cdot\color{blue}{\frac{1}{5}}\color{black}{}&\text{Simplify} \\ w=0&\text{Solution}\end{array}\nonumber\] Notice in Example \(\PageIndex{15}\) the seven canceled out completely. Many students get stuck at this point. Let’s not forget that the product of a number and zero is zero. Hence, zero is the solution to the equation. A common error students make with two-step equations is with negative signs. Remember, the sign always stays with the coefficient. Example \(\PageIndex{16}\)Solve for \(t\): \(4-2t=10\) \[\begin{array}{rl}4-2t=10&\text{Add the opposite of }4\text{ to each side} \\ 4-2t+\color{blue}{(-4)}\color{black}{}=10+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ -2t=6&\text{Multiply by the reciprocal of }-2 \\ \color{blue}{-\frac{1}{2}}\color{black}{}\cdot -2t=6\cdot\color{blue}{-\frac{1}{2}}\color{black}{}&\text{Simplify} \\ t=-3&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{17}\)Solve for \(n\): \(8 − n = 2\) \[\begin{array}{rl}8-n=2&\text{Add the opposite of }8\text{ to each side} \\ 8-n+\color{blue}{(-8)}\color{black}{}=2+\color{blue}{(-8)}\color{black}{}&\text{Simplify} \\ -n=-6&\text{Rewrite }-n\text{ as }-1n \\ -1n=-6&\text{Multiply by the reciprocal of }-1 \\ \color{blue}{-1}\color{black}{}\cdot -1n=-6\cdot\color{blue}{-1}\color{black}{}&\text{Simplify} \\ n=6&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{18}\)Solve for \(y\): \(−3y + 7 = −8\) \[\begin{array}{rl}-3y+7=-8&\text{Add the opposite of }7\text{ to each side} \\ -3y+7+\color{blue}{(-7)}\color{black}{}=-8+\color{blue}{(-7)}\color{black}{}&\text{Simplify} \\ -3y=-15&\text{Multiply by the reciprocal of }-3 \\ \color{blue}{-\frac{1}{3}}\color{black}{}\cdot -3y=-15\cdot\color{blue}{-\frac{1}{3}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{19}\)Solve for \(x\): \(−3 = \frac{x}{5} − 4\) Notice the variable term is on the right side of the equation. However, we still follow the method as usual: \[\begin{array}{rl}-3=\frac{x}{5}-4&\text{Add the opposite of }-4\text{ to each side} \\ -3+\color{blue}{4}\color{black}{}=\frac{x}{5}-4+\color{blue}{4}\color{black}{}&\text{Simplify} \\ 1=\frac{x}{5}&\text{Multiply by the reciprocal of }\frac{1}{5} \\ \color{blue}{5}\color{black}{}\cdot 1=\frac{x}{5}\cdot\color{blue}{5}\color{black}{}&\text{Simplify} \\ 5=x&\text{Rewrite the solution with the variable on the left side} \\ x=5&\text{Solution}\end{array}\nonumber\] As problems in algebra become more complex, the method remains the same. In fact, as we solve problems like those in the next example, each one of them will have several steps to solve, but the last two steps are a two-step equation. This is a critical reason to master two-step equations. Example \(\PageIndex{20}\)\[3x^2+4=y+6\qquad \frac{1}{x-8}+\frac{1}{x}=\frac{1}{3}\qquad \sqrt{5y-5}+1=y\qquad\log _5(2y-4)=1\nonumber\] Persian mathematician Omar Khayyam would solve algebraic problems geometrically by intersecting graphs rather than solving them algebraically. General EquationsOften as we are solving linear equations, we will need to do some preliminary work so that it is more familiar for us to solve. This section will focus on manipulating an equation in such a way that we can use our method for solving two-step equations to obtain the solution. One such issue that needs to be addressed is parentheses. Sometimes parenthesis can get in the way of solving an equation. As you might expect, we can rewrite an equation without parenthesis by using the distributive property. Distributive Property of EquationsThe following is an equivalent statement: \[c(a+b)=ca+cb\nonumber\] where \(c\neq 0\), i.e., we can multiply each term in the parenthesis by a nonzero number. Example \(\PageIndex{21}\)Solve for \(y\): \(4(2y − 6) = 16\) Notice the first step is distributing, then it is solved like any other two-step equation. \[\begin{array}{rl}4(2y-6)=16&\text{Distribute} \\ 8y-24=16&\text{Add the opposite of }-24\text{ to each side} \\ 8y-24+\color{blue}{24}\color{black}{}=16+\color{blue}{24}\color{black}{}&\text{Simplify} \\ 8y=40&\text{Multiply by the reciprocal of }8 \\ \color{blue}{\frac{1}{8}}\color{black}{}\cdot 8y=40\cdot\color{blue}{\frac{1}{8}}\color{black}{}&\text{Simplify} \\ y=5&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{22}\)Solve for \(p\): \(3(2p − 4) + 9 = 15\) \[\begin{array}{rl}3(2p-4)+9=15&\text{Distribute} \\ 6p-12+9=15&\text{Combine like terms} \\ 6p-3=15&\text{Add the opposite of }-3\text{ to each side} \\ 6p-3+\color{blue}{3}\color{black}{}=15+\color{blue}{3}\color{black}{}&\text{Simplify} \\ 6p=18&\text{Multiply by the reciprocal of }6 \\ \color{blue}{\frac{1}{6}}\color{black}{}\cdot 6p=18\cdot\color{blue}{\frac{1}{6}}\color{black}{}&\text{Simplify} \\ p=3&\text{Solution}\end{array}\nonumber\] Example \(\PageIndex{23}\)Solve for \(a\): \(3(4a − 5) = 4(2a + 1) + 5\) Notice we have variable terms on each side of the equation. We will not only distribute first, but also isolate the variable term before applying the two-step method: \[\begin{array}{rl} 3(4a − 5) = 4(2a + 1) + 5 &\text{Distribute} \\ 12a-15=8a+4+5&\text{Combine like terms} \\ 12a-15=8a+9&\text{Isolate the variable term by adding the opposite of }8a \\ &\text{to each side} \\ 12a-15+\color{blue}{(-8a)}\color{black}{}=8a+9+\color{blue}{(-8a)}\color{black}{}&\text{Simplify} \\ 4a-15=9&\text{Add the opposite of }-15\text{ to each side} \\ 4a-15+\color{blue}{15}\color{black}{}=9+\color{blue}{15}\color{black}{}&\text{Simplify} \\ 4a=24&\text{Multiply by the reciprocal of }4 \\ \color{blue}{\frac{1}{4}}\color{black}{}\cdot 4a=24\cdot\color{blue}{\frac{1}{4}}\color{black}{}&\text{Simplify} \\ a=6&\text{Solution}\end{array}\nonumber\] A general strategy to solving linear equationsIn general, there is a 5-step process to solving any linear equation. While all five steps aren’t always needed, this can serve as a guide for solving equations. Step 1. Apply the distributive property to rewrite the equation without parenthesis. Step 2. Combine like terms on each side of the equation. Step 3. Isolate the variable term by applying the addition property. Step 4. Solve the equation by applying the multiplication property. Step 5. \(✓\) Verify the solution. The Chinese developed a method for solving equations that involved finding each digit one at a time about 2,000 years ago. There are two special cases when solving linear equations. The first is illustrated in the next two examples. Notice we start by distributing and moving the variables all to the same side. Example \(\PageIndex{24}\)Solve for \(y\): \(3(2y − 5) = 6y − 15\) \[\begin{array}{rl} 3(2y − 5) = 6y − 15 &\text{Distribute} \\ 6y-15=6y-15&\text{Isolate the variable term by adding the opposite of }6y \\ &\text{to each side} \\ 6y-15+\color{blue}{(-6y)}\color{black}{}=6y+15+\color{blue}{(-6y)}\color{black}{}&\text{Simplify} \\ -15\stackrel{?}{=}-15&\text{Is this true? }\color{blue}{\text{Yes }\checkmark} \color{black}{} \\ -15=-15\end{array}\nonumber\] Notice all variables canceled and we are left with only a statement with numbers. In this case, the statement we are left with is a true statement, i.e., \(−15\) does equal \(−15\), and so there are infinitely many solutions to this equation. In this special case, when we obtain infinitely many solutions, then the solution is all real numbers . When the solution is all real numbers , we call this equation an identity . Example \(\PageIndex{25}\)Solve for \(n\): \(2(3n − 5) − 4n = 2n + 7\) \[\begin{array}{rl} 2(3n-5)-4n=2n+7&\text{Distribute} \\ 6n-10-4n=2n+7&\text{Combine like terms} \\ 2n-10=2n+7&\text{Isolate the variable term by adding the opposite of }2n \\ &\text{to each side} \\ -10\stackrel{?}{=}7 &\text{Is this true?}\color{blue}{\text{ No }X} \color{black}{} \\ -10\neq 7\end{array}\nonumber\] Notice all variables canceled and we are left with only a statement with numbers. In this case, the statement we are left with is a false statement, i.e., \(−10\) does not equal \(7\), and so there is no solution to this equation. In this special case, when we obtain a false statement, then the solution is no solution and we call this equation a contradiction . Conditional, Identity and Contradiction EquationsThere are three types of equations we saw in the examples and in Examples \(\PageIndex{24}\) and \(\PageIndex{25}\) :
Solving Equations with FractionsOften when solving linear equations, we work with an equation with fraction coefficients. We can solve these problems as usual. Recall, the methods never change, just problems. Example \(\PageIndex{26}\)Solve for \(y\): \(\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}\) \[\begin{array}{rl}\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}&\text{Isolate the variable term by adding the opposite of }-\frac{7}{2} \\ \frac{3}{4}y-\frac{7}{2}+\color{blue}{\frac{7}{2}}\color{black}{}=\frac{5}{6}+\color{blue}{\frac{7}{2}}\color{black}{}&\text{Simplify}\end{array}\nonumber\] Notice, in order to add \(\frac{5}{6} + \frac{7}{2}\), we need to obtain the LCD. Since the \(\text{LCD}(2, 6) = 6\), we can rewrite the right side with the LCD: \[\begin{array}{rl}\frac{3}{4}y=\frac{5}{6}+\frac{21}{6}&\text{Combine like terms} \\ \frac{3}{4}y=\frac{26}{6}&\text{Simplify }\frac{26}{6} \\ \frac{3}{4}y=\frac{13}{3}&\text{Solve by multiplying by the reciprocal of }\frac{3}{4} \\ \color{blue}{\frac{4}{3}}\color{black}{}\cdot\frac{3}{4}y=\frac{13}{3}\cdot\color{blue}{\frac{4}{3}}\color{black}{}&\text{Simplify} \\ y=\frac{52}{9}&\text{Solution}\end{array}\nonumber\] While this process does help us arrive at the correct solution, the fractions can make the process quite difficult and we are more inclined to make errors. Hence, we have an alternate method called clearing denominators . Clearing DenominatorsWe can easily clear denominators in an equation by multiplying each term by the LCD. After completing this step, the fractions are cleared and we can work with a more familiar type of equation. Let’s try Example \(\PageIndex{26}\) again, but, now, by clearing denominators first, then solving. Example \(\PageIndex{27}\)\[\begin{array}{rl}\frac{3}{4}y-\frac{7}{2}=\frac{5}{6}&\text{Multiply each term by the LCD}(2,4,6)=12 \\ \color{blue}{12}\color{black}{}\cdot\frac{3}{4}y-\color{blue}{12}\color{black}{}\cdot\frac{7}{2}=\color{blue}{12}\color{black}{}\cdot\frac{5}{6}&\text{Simplify} \\ \frac{36}{4}y-\frac{84}{2}=\frac{60}{6}&\text{Reduce the fractions} \\ 9y-42=10&\text{Add the opposite of }-42\text{ to each side} \\ 9y-42+\color{blue}{42}\color{black}{}=10+\color{blue}{42}\color{black}{}&\text{Simplify} \\ 9y=52&\text{Multiply by the reciprocal of }9 \\ \color{blue}{\frac{1}{9}}\color{black}{}\cdot 9y=52\cdot\color{blue}{\frac{1}{9}}\color{black}{}&\text{Simplify} \\ y=\frac{52}{9}&\text{Solution}\end{array}\nonumber\] Thus, the solution is \(y = \frac{52}{9}\) and Example \(\PageIndex{26}\) is a conditional equation . Also, we observe that as soon as we multiplied each term by the LCD, we cleared the denominators and the equation no longer contained fractions. Example \(\PageIndex{28}\)Solve for \(t\): \(\frac{2}{3}t-2=\frac{3}{2}t+\frac{1}{6}\) \[\begin{array}{rl}\frac{2}{3}t-2=\frac{3}{2}t+\frac{1}{6}&\text{Multiply each term by the LCD}(2,3,6)=6 \\ \color{blue}{6}\color{black}{}\cdot\frac{2}{3}t-\color{blue}{6}\color{black}{}\cdot 2=\color{blue}{6}\color{black}{}\cdot\frac{3}{2}t+\color{blue}{6}\color{black}{}\cdot\frac{1}{6}&\text{Simplify} \\ \frac{12}{3}t-12=\frac{18}{2}t+\frac{6}{6}&\text{Reduce the fractions} \\ 4t-12=9t+1&\text{Isolate the variable term by adding the opposite of }9t \\ &\text{to each side} \\ 4t-12+\color{blue}{(-9t)}\color{black}{}=9t+1+\color{blue}{(-9t)}\color{black}{}&\text{Simplify} \\ -5t-12=1&\text{Add the opposite of }-12\text{ to each side} \\ -5t-12+\color{blue}{12}\color{black}{}=1+\color{blue}{12}\color{black}{}&\text{Simplify} \\ -5t=13&\text{Multiply by the reciprocal of }-5 \\ \color{blue}{-\frac{1}{5}}\color{black}{}\cdot -5t=13\cdot\color{blue}{-\frac{1}{5}}\color{black}{}&\text{Simplify} \\ t=-\frac{13}{5}&\text{Solution}\end{array}\nonumber\] Thus, the solution is \(t = − \frac{13}{5}\) and Example \(\PageIndex{28}\) is a conditional equation . In Example \(\PageIndex{28}\) , we could write the solution as \(−2.6\) given we were using a calculator. A good rule of thumb is if you start with fractions, the solution should also be a fraction. Hence, since there were fractions in original problem, we will leave the solution as a fraction. The Egyptians were among the first to study fractions and linear equations. The most famous mathematical document from Ancient Egypt is the Rhind Papyrus , where the unknown variable was called “heap.” Solving Equations with Distributing FractionsWe can use this same method if parenthesis are in the given problem. We will first distribute the coefficient in front of the parenthesis, then clear denominators. Example \(\PageIndex{29}\)Solve for \(y\): \(\frac{3}{2}\left(\frac{5}{9}y+\frac{4}{27}\right)=3\) \[\begin{array}{rl}\frac{3}{2}\left(\frac{5}{9}y+\frac{4}{27}\right)=3&\text{Distribute }\frac{3}{2}\text{ and reduce} \\ \frac{5}{6}y+\frac{2}{9}=3&\text{Multiply each term by the LCD}(6,9)=18 \\ \color{blue}{18}\color{black}{}\cdot\frac{5}{6}y+\color{blue}{18}\color{black}{}\cdot\frac{2}{9}=\color{blue}{18}\color{black}{}\cdot 3&\text{Multiply and simplify} \\ 15y+4=54&\text{Add the opposite of }4\text{ to each side} \\ 15y+4+\color{blue}{(-4)}\color{black}{}=54+\color{blue}{(-4)}\color{black}{}&\text{Simplify} \\ 15y=50&\text{Multiply by the reciprocal of }15 \\ \color{blue}{\frac{1}{15}}\color{black}{}\cdot 15y=50\cdot\color{blue}{\frac{1}{15}}\color{black}{}&\text{Simplify} \\ y=\frac{50}{15}&\text{Reduce} \\ y=\frac{10}{3}&\text{Solution}\end{array}\nonumber\] Thus, the solution is \(y = \frac{10}{3}\) and Example \(\PageIndex{29}\) is a conditional equation . Example \(\PageIndex{30}\)Solve for \(q\): \(\frac{1}{4}q-\frac{1}{2}=\frac{1}{3}\left(\frac{3}{4}q+6\right)-\frac{7}{2}\) \[\begin{array}{rl}\frac{1}{4}q-\frac{1}{2}=\frac{1}{3}\left(\frac{3}{4}q+6\right)-\frac{7}{2}&\text{Distribute }\frac{1}{3}\text{ and reduce} \\ \frac{1}{4}q-\frac{1}{2}=\frac{1}{4}q+2-\frac{7}{2}&\text{Multiply each term by the LCD}(2,4)=4 \\ \color{blue}{4}\color{black}{}\cdot\frac{1}{4}q-\color{blue}{4}\color{black}{}\cdot\frac{1}{2}=\color{blue}{4}\color{black}{}\cdot\frac{1}{4}q+\color{blue}{4}\color{black}{}\cdot 2-\color{blue}{4}\color{black}{}\cdot\frac{7}{2}&\text{Multiply and reduce} \\ q-2=q+8-14&\text{Combine like terms} \\ q-2=q-6&\text{Isolate the variable term by adding the opposite of }q \\ &\text{to each side} \\ q-2+\color{blue}{(-q)}\color{black}{}=q-6+\color{blue}{(-q)}\color{black}{}&\text{Simplify} \\ -2\stackrel{?}{=}-6&\text{Is this true? }\color{blue}{\text{No }X}\color{black}{} \\ -2\neq -6&\text{This implies there is no solution}\end{array}\nonumber\] Since we obtain a false statement, there is no solution and this equation is called a contradiction . Linear Equations HomeworkSolve the one-step equations. Exercise \(\PageIndex{1}\)\(v + 9 = 16\) Exercise \(\PageIndex{2}\)\(x − 11 = −16\) Exercise \(\PageIndex{3}\)\(30 = a + 20\) Exercise \(\PageIndex{4}\)\(x − 7 = −26\) Exercise \(\PageIndex{5}\)\(13 = n − 5\) Exercise \(\PageIndex{6}\)\(340 = −17x\) Exercise \(\PageIndex{7}\)\(−9 = \frac{n}{12}\) Exercise \(\PageIndex{8}\)\(20v = −160\) Exercise \(\PageIndex{9}\)\(340 = 20n\) Exercise \(\PageIndex{10}\)\(16x = 320\) Exercise \(\PageIndex{11}\)\(−16 + n = −13\) Exercise \(\PageIndex{12}\)\(p − 8 = −21\) Exercise \(\PageIndex{13}\)\(180 = 12x\) Exercise \(\PageIndex{14}\)\(20b = −200\) Exercise \(\PageIndex{15}\)\(\frac{r}{14}=\frac{5}{14}\) Exercise \(\PageIndex{16}\)\(−7 = a + 4\) Exercise \(\PageIndex{17}\)\(10 = x − 4\) Exercise \(\PageIndex{18}\)\(13a = −143\) Exercise \(\PageIndex{19}\)\(\frac{p}{20} = −12\) Exercise \(\PageIndex{20}\)\(9 + m = −7\) Exercise \(\PageIndex{21}\)\(14 = b + 3\) Exercise \(\PageIndex{22}\)\(−14 = x − 18\) Exercise \(\PageIndex{23}\)\(−1 + k = 5\) Exercise \(\PageIndex{24}\)\(−13 + p = −19\) Exercise \(\PageIndex{25}\)\(22 = 16 + m\) Exercise \(\PageIndex{26}\)\(4r = −28\) Exercise \(\PageIndex{27}\)\(\frac{5}{9} = \frac{b}{9}\) Exercise \(\PageIndex{28}\)\(−20x = −80\) Exercise \(\PageIndex{29}\)\(\frac{1}{2} = \frac{a}{8}\) Exercise \(\PageIndex{30}\)\(\frac{k}{13}= −16\) Exercise \(\PageIndex{31}\)\(21 = x + 5\) Exercise \(\PageIndex{32}\)\(m − 4 = −13\) Exercise \(\PageIndex{33}\)\(3n = 24\) Exercise \(\PageIndex{34}\)\(−17 = \frac{x}{12}\) Exercise \(\PageIndex{35}\)\(n + 8 = 10\) Exercise \(\PageIndex{36}\)\(v − 16 = −30\) Exercise \(\PageIndex{37}\)\(−15 = x − 16\) Exercise \(\PageIndex{38}\)\(-8k=120\) Exercise \(\PageIndex{39}\)\(-15=\frac{x}{9}\) Exercise \(\PageIndex{40}\)\(-19=\frac{n}{20}\) Solve the two-step equations. Exercise \(\PageIndex{41}\)\(5 + \frac{n}{4}= 4\) Exercise \(\PageIndex{42}\)\(102 = −7r + 4\) Exercise \(\PageIndex{43}\)\(−8n + 3 = −77\) Exercise \(\PageIndex{44}\)\(0 = −6v\) Exercise \(\PageIndex{45}\)\(−8 = \frac{x}{5}− 6\) Exercise \(\PageIndex{46}\)\(0 = −7 + \frac{k}{2}\) Exercise \(\PageIndex{47}\)\(−12 + 3x = 0\) Exercise \(\PageIndex{48}\)\(24 = 2n − 8\) Exercise \(\PageIndex{49}\)\(2 = −12 + 2r\) Exercise \(\PageIndex{50}\)\(\frac{b}{3} + 7 = 10\) Exercise \(\PageIndex{51}\)\(152 = 8n + 64\) Exercise \(\PageIndex{52}\)\(−16 = 8a + 64\) Exercise \(\PageIndex{53}\)\(56 + 8k = 64\) Exercise \(\PageIndex{54}\)\(−2x + 4 = 22\) Exercise \(\PageIndex{55}\)\(−20 = 4p + 4\) Exercise \(\PageIndex{56}\)\(−5 = 3 + \frac{n}{2}\) Exercise \(\PageIndex{57}\)\(\frac{r}{8} − 6 = −5\) Exercise \(\PageIndex{58}\)\(−40 = 4n − 32\) Exercise \(\PageIndex{59}\)\(87 = 3 − 7v\) Exercise \(\PageIndex{60}\)\(−x + 1 = −11\) Exercise \(\PageIndex{61}\)\(−2 = −2m + 12\) Exercise \(\PageIndex{62}\)\(27 = 21 − 3x\) Exercise \(\PageIndex{63}\)\(−4 − b = 8\) Exercise \(\PageIndex{64}\)\(−2 + \frac{x}{2} = 4\) Exercise \(\PageIndex{65}\)\(−5 = \frac{a}{4} − 1\) Exercise \(\PageIndex{66}\)\(−6 = 15 + 3p\) Exercise \(\PageIndex{67}\)\(−5m + 2 = 27\) Exercise \(\PageIndex{68}\)\(−37 = 8 + 3x\) Exercise \(\PageIndex{69}\)\(−8 + \frac{n}{12} = −7\) Exercise \(\PageIndex{70}\)\(\frac{x}{1} − 8 = −8\) Exercise \(\PageIndex{71}\)\(−11 = −8 + \frac{v}{2}\) Exercise \(\PageIndex{72}\)\(−2x − 3 = −29\) Exercise \(\PageIndex{73}\)\(−4 − 3n = −16\) Exercise \(\PageIndex{74}\)\(67 = 5m − 8\) Exercise \(\PageIndex{75}\)\(9 = 8 + \frac{x}{6}\) Exercise \(\PageIndex{76}\)\(\frac{m}{4} − 1 = −2\) Exercise \(\PageIndex{77}\)\(−80 = 4x − 28\) Exercise \(\PageIndex{78}\)\(33 = 3b + 3\) Exercise \(\PageIndex{79}\)\(3x − 3 = −3\) Exercise \(\PageIndex{80}\)\(4 + \frac{a}{3} = 1\) Exercise \(\PageIndex{81}\)\(2 − (−3a − 8) = 1\) Exercise \(\PageIndex{82}\)\(−5 (−4 + 2v) = −50\) Exercise \(\PageIndex{83}\)\(66 = 6 (6 + 5x)\) Exercise \(\PageIndex{84}\)\(0 = −8 (p − 5)\) Exercise \(\PageIndex{85}\)\(−2 + 2 (8x − 7) = −16\) Exercise \(\PageIndex{86}\)\(−21x + 12 = −6 − 3x\) Exercise \(\PageIndex{87}\)\(−1 − 7m = −8m + 7\) Exercise \(\PageIndex{88}\)\(1 − 12r = 29 − 8r\) Exercise \(\PageIndex{89}\)\(20 − 7b = −12b + 30\) Exercise \(\PageIndex{90}\)\(−32 − 24v = 34 − 2v\) Exercise \(\PageIndex{91}\)\(−2 − 5 (2 − 4m) = 33 + 5m\) Exercise \(\PageIndex{92}\)\(−4n + 11 = 2 (1 − 8n) + 3n\) Exercise \(\PageIndex{93}\)\(−6v − 29 = −4v − 5 (v + 1)\) Exercise \(\PageIndex{94}\)\(2 (4x − 4) = −20 − 4x\) Exercise \(\PageIndex{95}\)\(−a − 5 (8a − 1) = 39 − 7a\) Exercise \(\PageIndex{96}\)\(−57 = − (−p + 1) + 2 (6 + 8p)\) Exercise \(\PageIndex{97}\)\(−2 (m − 2) + 7 (m − 8) = −67\) Exercise \(\PageIndex{98}\)\(50 = 8 (7 + 7r) − (4r + 6)\) Exercise \(\PageIndex{99}\)\(−8 (n − 7) + 3 (3n − 3) = 41\) Exercise \(\PageIndex{100}\)\(−61 = −5 (5r − 4) + 4 (3r − 4)\) Exercise \(\PageIndex{101}\)\(−2 (8n − 4) = 8 (1 − n)\) Exercise \(\PageIndex{102}\)\(−3 (−7v + 3) + 8v = 5v − 4 (1 − 6v)\) Exercise \(\PageIndex{103}\)\(−7 (x − 2) = −4 − 6 (x − 1)\) Exercise \(\PageIndex{104}\)\(−6 (8k + 4) = −8 (6k + 3) − 2\) Exercise \(\PageIndex{105}\)\(−2 (1 − 7p) = 8 (p − 7)\) Exercise \(\PageIndex{106}\)\(2 (−3n + 8) = −20\) Exercise \(\PageIndex{107}\)\(2 − 8 (−4 + 3x) = 34\) Exercise \(\PageIndex{108}\)\(32 = 2 − 5 (−4n + 6)\) Exercise \(\PageIndex{109}\)\(−55 = 8 + 7 (k − 5)\) Exercise \(\PageIndex{110}\)\(− (3 − 5n) = 12\) Exercise \(\PageIndex{111}\)\(−3n − 27 = −27 − 3n\) Exercise \(\PageIndex{112}\)\(56p − 48 = 6p + 2\) Exercise \(\PageIndex{113}\)\(4 + 3x = −12x + 4\) Exercise \(\PageIndex{114}\)\(−16n + 12 = 39 − 7n\) Exercise \(\PageIndex{115}\)\(17 − 2x = 35 − 8x\) Exercise \(\PageIndex{116}\)\(−25 − 7x = 6 (2x − 1)\) Exercise \(\PageIndex{117}\)\(−7 (1 + b) = −5 − 5b\) Exercise \(\PageIndex{118}\)\(−8 (8r − 2) = 3r + 16\) Exercise \(\PageIndex{119}\)\(−8n − 19 = −2 (8n − 3) + 3n\) Exercise \(\PageIndex{120}\)\(−4 + 4k = 4 (8k − 8)\) Exercise \(\PageIndex{121}\)\(16 = −5 (1 − 6x) + 3 (6x + 7)\) Exercise \(\PageIndex{122}\)\(7 = 4 (n − 7) + 5 (7n + 7)\) Exercise \(\PageIndex{123}\)\(−8 (6 + 6x) + 4 (−3 + 6x) = −12\) Exercise \(\PageIndex{124}\)\(−76 = 5 (1 + 3b) + 3 (3b − 3)\) Exercise \(\PageIndex{125}\)\(−6 (x − 8) − 4 (x − 2) = −4\) Exercise \(\PageIndex{126}\)\(−4 (1 + a) = 2a − 8 (5 + 3a)\) Exercise \(\PageIndex{127}\)\(−6 (x − 3) + 5 = −2 − 5 (x − 5)\) Exercise \(\PageIndex{128}\)\(− (n + 8) + n = −8n + 2 (4n − 4)\) Exercise \(\PageIndex{129}\)\(−5 (x + 7) = 4 (−8x − 2)\) Exercise \(\PageIndex{130}\)\(8 (−8n + 4) = 4 (−7n + 8)\) Exercise \(\PageIndex{131}\)\(\frac{3}{5}(1+p)=\frac{21}{20}\) Exercise \(\PageIndex{132}\)\(0=-\frac{5}{4}\left(x-\frac{6}{5}\right)\) Exercise \(\PageIndex{133}\)\(\frac{3}{4}-\frac{5}{4}m=\frac{113}{24}\) Exercise \(\PageIndex{134}\)\(\frac{635}{72}=-\frac{5}{2}\left(-\frac{11}{4}+x\right)\) Exercise \(\PageIndex{135}\)\(2b+\frac{9}{5}=-\frac{11}{5}\) Exercise \(\PageIndex{136}\)\(\frac{3}{2}\left(\frac{7}{3}n+1\right)=\frac{3}{2}\) Exercise \(\PageIndex{137}\)\(-a-\frac{5}{4}\left(-\frac{8}{3}a+1\right)=-\frac{19}{4}\) Exercise \(\PageIndex{138}\)\(\frac{55}{6}=-\frac{5}{2}\left(\frac{3}{2}p-\frac{5}{3}\right)\) Exercise \(\PageIndex{139}\)\(\frac{16}{9}=-\frac{4}{3}\left(-\frac{4}{3}n-\frac{4}{3}\right)\) Exercise \(\PageIndex{140}\)\(-\frac{5}{8}=\frac{5}{4}\left(r-\frac{3}{2}\right)\) Exercise \(\PageIndex{141}\)\(-\frac{11}{3}+\frac{3}{2}b=\frac{5}{2}\left(b-\frac{5}{3}\right)\) Exercise \(\PageIndex{142}\)\(-\left(-\frac{5}{2}x-\frac{3}{2}\right)=-\frac{3}{2}+x\) Exercise \(\PageIndex{143}\)\(\frac{45}{16}+\frac{3}{2}n=\frac{7}{4}n-\frac{19}{16}\) Exercise \(\PageIndex{144}\)\(\frac{3}{2}\left(v+\frac{3}{2}\right)=-\frac{7}{4}v-\frac{19}{6}\) Exercise \(\PageIndex{145}\)\(\frac{47}{9}+\frac{3}{2}x=\frac{5}{3}\left(\frac{5}{2}x+1\right)\) Exercise \(\PageIndex{146}\)\(-\frac{1}{2}=\frac{3}{2}k+\frac{3}{2}\) Exercise \(\PageIndex{147}\)\(\frac{3}{2}n-\frac{8}{3}=-\frac{29}{12}\) Exercise \(\PageIndex{148}\)\(\frac{11}{4}+\frac{3}{4}r=\frac{163}{32}\) Exercise \(\PageIndex{149}\)\(-\frac{16}{9}=-\frac{4}{3}\left(\frac{5}{3}+n\right)\) Exercise \(\PageIndex{150}\)\(\frac{3}{2}-\frac{7}{4}v=-\frac{9}{8}\) Exercise \(\PageIndex{151}\)\(\frac{41}{9}=\frac{5}{2}\left(x+\frac{2}{3}\right)-\frac{1}{3}x\) Exercise \(\PageIndex{152}\)\(\frac{1}{3}\left(-\frac{7}{4}k+1\right)-\frac{10}{3}k=-\frac{13}{8}\) Exercise \(\PageIndex{153}\)\(-\frac{1}{2}\left(\frac{2}{3}x-\frac{3}{4}\right)-\frac{7}{2}x=-\frac{83}{24}\) Exercise \(\PageIndex{154}\)\(\frac{2}{3}\left(m+\frac{9}{4}\right)-\frac{10}{3}=-\frac{53}{18}\) Exercise \(\PageIndex{155}\)\(\frac{1}{12}=\frac{4}{3}x+\frac{5}{3}\left(x-\frac{7}{4}\right)\) Exercise \(\PageIndex{156}\)\(\frac{7}{6}-\frac{4}{3}n=-\frac{3}{2}n+2\left(n+\frac{3}{2}\right)\) Exercise \(\PageIndex{157}\)\(-\frac{149}{16}-\frac{11}{3}r=-\frac{7}{4}r-\frac{5}{4}\left(-\frac{4}{3}r+1\right)\) Exercise \(\PageIndex{158}\)\(-\frac{7}{2}\left(\frac{5}{3}a+\frac{1}{3}\right)=\frac{11}{4}a+\frac{25}{8}\) Exercise \(\PageIndex{159}\)\(-\frac{8}{3}-\frac{1}{2}x=-\frac{4}{3}x-\frac{2}{3}\left(-\frac{13}{4}x+1\right)\) Exercise \(\PageIndex{160}\)\(\frac{1}{3}n+\frac{29}{6}=2\left(\frac{4}{3}n+\frac{2}{3}\right)\) If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Want additional help with Algebra 2? Unit 1: Polynomial arithmeticUnit 2: complex numbers, unit 3: polynomial factorization, unit 4: polynomial division, unit 5: polynomial graphs, unit 6: rational exponents and radicals, unit 7: exponential models, unit 8: logarithms, unit 9: transformations of functions, unit 10: equations, unit 11: trigonometry, unit 12: modeling. |
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This topic covers: - Intercepts of linear equations/functions - Slope of linear equations/functions - Slope-intercept, point-slope, & standard forms - Graphing linear equations/functions - Writing linear equations/functions - Interpreting linear equations/functions - Linear equations/functions word problems
Solving Linear Equations in One Variable. A linear equation is an equation of a straight line, written in one variable. The only power of the variable is 1. Linear equations in one variable may take the form a x + b = 0 a x + b = 0 and are solved using basic algebraic operations. We begin by classifying linear equations in one variable as one ...
A linear equation of the form takes two steps to solve. First, use the appropriate equality property of addition or subtraction to isolate the variable term. Next, isolate the variable using the equality property of multiplication or division. Checking solutions in the following examples is left to the reader.
SECTION 2.1 Linear Equations MATH 1310 College Algebra 83 Solution: Additional Example 1: Solution: CHAPTER 2 Solving Equations and Inequalities 84 University of Houston Department of Mathematics Additional Example 2: Solution: Additional Example 3: Solution: We first multiply both sides of the equation by 12 to clear the equation of fractions. ...
y = −2x + 5 y = − 2 x + 5. This page titled 2.4: Graphing Linear Equations- Answers to the Homework Exercises is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Darlene Diaz ( ASCCC Open Educational Resources Initiative) via source content that was edited to the style and standards of the LibreTexts ...
Assuming the relationship between the female students' weights and heights is linear, write an equation giving the relationship between heights and weights of female students, and use this relationship to predict the weight of a female student who is 70 inches tall. Rylie Howey. Numerade Educator. 02:34.
Good question! In x and/or y, any linear equation is equivalent to one of two forms: x=a or y=mx+b where a, m, and b are constants. (Yes, this already includes the form where y is a constant, because this would be the result of taking m to be 0 in the equation y=mx+b).
2.2: Use a General Strategy to Solve Linear Equations. Solving an equation is like discovering the answer to a puzzle. The purpose in solving an equation is to find the value or values of the variable that makes it a true statement. Any value of the variable that makes the equation true is called a solution to the equation.
3.4 Graphing Linear Equations. There are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation. If the equation is given in the form y = mx+b y = m x + b, then m m gives the rise over run value and ...
Functions and linear models. Unit 15. Systems of equations. Course challenge. Test your knowledge of the skills in this course. Start Course challenge. Math; ... Two-step equations with decimals and fractions Get 5 of 7 questions to level up! Find the mistake: two-step equations Get 3 of 4 questions to level up!
Find here an unlimited supply of printable worksheets for solving linear equations, available as both PDF and html files. You can customize the worksheets to include one-step, two-step, or multi-step equations, variable on both sides, parenthesis, and more. The worksheets suit pre-algebra and algebra 1 courses (grades 6-9).
Lesson 7. Systems of Linear Equations (Primarily 3 by 3) LESSON/HOMEWORK. LESSON VIDEO. ANSWER KEY. EDITABLE LESSON. EDITABLE KEY.
This page titled Chapter 2: Linear Equations is shared under a CK-12 license and was authored, remixed, and/or curated by CK-12 Foundation via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
Homework 1: Linear Equations This homework is due on Monday, January 29, respectively Tuesday January 30, 2018. Homework is due at the beginning of each class in the classroom. 1 Find all solutions of the linear system x+ y + z + u = 1 x+ y u+ v = 2 x+ z = 3 x+ y + u = 4 y + v = 5 Solution: x = 6;y = 11;z = 3;u = 9;v = 16. 2 On the iWatch one ...
The Algebra 1 course, often taught in the 9th grade, covers Linear equations, inequalities, functions, and graphs; Systems of equations and inequalities; Extension of the concept of a function; Exponential models; and Quadratic equations, functions, and graphs. Khan Academy's Algebra 1 course is built to deliver a comprehensive, illuminating, engaging, and Common Core aligned experience!
2.2 Solving Linear Equations. When working with questions that require two or more steps to solve, do the reverse of the order of operations to solve for the variable. Example 2.2.1. Solve 4x +16 = −4 4 x + 16 = − 4 for x. x. 4x + 16 = −4 −16 −16 Subtract 16 from each side 4x 4 = −20 4 Divide each side by 4 x = −5 Solution 4 x ...
Solve an algebraic equation using the addition property of equality. First, let's define some important terminology: variables: variables are symbols that stand for an unknown quantity, they are often represented with letters, like x, y, or z. coefficient: Sometimes a variable is multiplied by a number.This number is called the coefficient of the variable.
Unit 4: Linear Equations Homework 10: Parallel & Perpendicular Lines (Day 2) Write an equation passing through the point and PARALLEL to the given line. + 6 5.1 = +15 6. (-5, -1); 2x-4 5. (-10, 1); 21 + 9 = 15 Directions: Write an equation passing through the point and PERPENDICULAR to the given line. +10 11. (10, 7); 5x-6y= 18
Math 1111 College Algebra: Secion 1 Linear Equaions and Raional Equaion. I. Solving Linear Equaions in One Variable Example 1: Solving a Linear Equaion Solve and check: 2x + 3 = 17 Example 2: Solving a Linear Equaion Solve and check: 2(x - 3) - 17 = 13 - 3(x + 2) II. Linear Equaions with Fracions Example 3: Solving a Linear Equaion ...
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A linear equation is an equation where the highest exponent on the given variables is one. A linear equation in one variable is an equation with one variable with exponent one, e.g., ax + b = c, where a is called the coefficient of x, and b and c are constant coefficients. Solving linear equations is an important and fundamental skill in algebra.
Modeling with Linear Equations. SOLVING AN APPLIED PROBLEM. Step 1 Read the problem carefully until you understand what is given and what is to be found. Step 2 Assign a variable to represent the unknown value, using diagrams or tables as needed. Write down what the variable represents. If necessary, express any other unknown values in terms of ...
The Algebra 2 course, often taught in the 11th grade, covers Polynomials; Complex Numbers; Rational Exponents; Exponential and Logarithmic Functions; Trigonometric Functions; Transformations of Functions; Rational Functions; and continuing the work with Equations and Modeling from previous grades. Khan Academy's Algebra 2 course is built to deliver a comprehensive, illuminating, engaging, and ...