Skip to Content
OpenStax Logo
College Algebra

2.2 Linear Equations in One Variable

College Algebra2.2 Linear Equations in One Variable
  1. Preface
  2. 1 Prerequisites
    1. Introduction to Prerequisites
    2. 1.1 Real Numbers: Algebra Essentials
    3. 1.2 Exponents and Scientific Notation
    4. 1.3 Radicals and Rational Exponents
    5. 1.4 Polynomials
    6. 1.5 Factoring Polynomials
    7. 1.6 Rational Expressions
    8. Key Terms
    9. Key Equations
    10. Key Concepts
    11. Review Exercises
    12. Practice Test
  3. 2 Equations and Inequalities
    1. Introduction to Equations and Inequalities
    2. 2.1 The Rectangular Coordinate Systems and Graphs
    3. 2.2 Linear Equations in One Variable
    4. 2.3 Models and Applications
    5. 2.4 Complex Numbers
    6. 2.5 Quadratic Equations
    7. 2.6 Other Types of Equations
    8. 2.7 Linear Inequalities and Absolute Value Inequalities
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  4. 3 Functions
    1. Introduction to Functions
    2. 3.1 Functions and Function Notation
    3. 3.2 Domain and Range
    4. 3.3 Rates of Change and Behavior of Graphs
    5. 3.4 Composition of Functions
    6. 3.5 Transformation of Functions
    7. 3.6 Absolute Value Functions
    8. 3.7 Inverse Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  5. 4 Linear Functions
    1. Introduction to Linear Functions
    2. 4.1 Linear Functions
    3. 4.2 Modeling with Linear Functions
    4. 4.3 Fitting Linear Models to Data
    5. Key Terms
    6. Key Concepts
    7. Review Exercises
    8. Practice Test
  6. 5 Polynomial and Rational Functions
    1. Introduction to Polynomial and Rational Functions
    2. 5.1 Quadratic Functions
    3. 5.2 Power Functions and Polynomial Functions
    4. 5.3 Graphs of Polynomial Functions
    5. 5.4 Dividing Polynomials
    6. 5.5 Zeros of Polynomial Functions
    7. 5.6 Rational Functions
    8. 5.7 Inverses and Radical Functions
    9. 5.8 Modeling Using Variation
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  7. 6 Exponential and Logarithmic Functions
    1. Introduction to Exponential and Logarithmic Functions
    2. 6.1 Exponential Functions
    3. 6.2 Graphs of Exponential Functions
    4. 6.3 Logarithmic Functions
    5. 6.4 Graphs of Logarithmic Functions
    6. 6.5 Logarithmic Properties
    7. 6.6 Exponential and Logarithmic Equations
    8. 6.7 Exponential and Logarithmic Models
    9. 6.8 Fitting Exponential Models to Data
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  8. 7 Systems of Equations and Inequalities
    1. Introduction to Systems of Equations and Inequalities
    2. 7.1 Systems of Linear Equations: Two Variables
    3. 7.2 Systems of Linear Equations: Three Variables
    4. 7.3 Systems of Nonlinear Equations and Inequalities: Two Variables
    5. 7.4 Partial Fractions
    6. 7.5 Matrices and Matrix Operations
    7. 7.6 Solving Systems with Gaussian Elimination
    8. 7.7 Solving Systems with Inverses
    9. 7.8 Solving Systems with Cramer's Rule
    10. Key Terms
    11. Key Equations
    12. Key Concepts
    13. Review Exercises
    14. Practice Test
  9. 8 Analytic Geometry
    1. Introduction to Analytic Geometry
    2. 8.1 The Ellipse
    3. 8.2 The Hyperbola
    4. 8.3 The Parabola
    5. 8.4 Rotation of Axes
    6. 8.5 Conic Sections in Polar Coordinates
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Review Exercises
    11. Practice Test
  10. 9 Sequences, Probability, and Counting Theory
    1. Introduction to Sequences, Probability and Counting Theory
    2. 9.1 Sequences and Their Notations
    3. 9.2 Arithmetic Sequences
    4. 9.3 Geometric Sequences
    5. 9.4 Series and Their Notations
    6. 9.5 Counting Principles
    7. 9.6 Binomial Theorem
    8. 9.7 Probability
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Review Exercises
    13. Practice Test
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
  12. Index

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.

Coordinate plane where the x-axis ranges from 0 to 200 in intervals of 20 and the y-axis ranges from 0 to 3,000 in intervals of 500.  The x-axis is labeled Hours Worked and the y-axis is labeled Savings Account Balance.  A linear function is plotted with a y-intercept of 400 with a slope of 15.  A dotted horizontal line extends from the point (0,2500).
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 ax+b=0 ax+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.

3x=2x+x 3x=2x+x

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 5x+2=3x6, 5x+2=3x6,we have the following:

5x+2 = 3x6 2x = −8 x = −4 5x+2 = 3x6 2x = −8 x = −4

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 5x15=5( x4 ), 5x15=5( x4 ),we have the following:

5x15 = 5x20 5x155x = 5x205x Subtract 5x from both sides. −15 −20 False statement 5x15 = 5x20 5x155x = 5x205x Subtract 5x from both sides. −15 −20 False statement

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

ax+b=0 ax+b=0

where a and b are real numbers, a0. a0.

How To

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:

  1. 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.
  2. Apply the distributive property as needed: a( b+c )=ab+ac. a( b+c )=ab+ac.
  3. Isolate the variable on one side of the equation.
  4. When the variable is multiplied by a coefficient in the final stage, multiply both sides of the equation by the reciprocal of the coefficient.

Example 1

Solving an Equation in One Variable

Solve the following equation: 2x+7=19. 2x+7=19.

Try It #1

Solve the linear equation in one variable: 2x+1=−9. 2x+1=−9.

Example 2

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 ).

Analysis

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 .

Try It #2

Solve the equation in one variable: −2( 3x1 )+x=14x. −2( 3x1 )+x=14x.

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.

x+1 x 2 4 , 1 x3 ,or 4 x 2 +x2 x+1 x 2 4 , 1 x3 ,or 4 x 2 +x2

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.

Example 3

Solving a Rational Equation

Solve the rational equation: 7 2x 5 3x = 22 3 . 7 2x 5 3x = 22 3 .

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, x1, x1,and 3x3. 3x3.First, factor all denominators. We then have x, x, (x1), (x1),and 3(x1) 3(x1)as the denominators. (Note the parentheses placed around the second denominator.) Only the last two denominators have a common factor of (x1). (x1).The x x in the first denominator is separate from the x x in the (x1) (x1)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 ( x1 ), ( x1 ), and the 3. Thus, the LCD is the following:

x( x1 )3=3x( x1 ) x( x1 )3=3x( x1 )

So, both sides of the equation would be multiplied by 3x( x1 ). 3x( x1 ).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 +2x. x 2 +2x. Once the second denominator is factored as x 2 +2x=x( x+2 ), x 2 +2x=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.

a b = c d a b = c d

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 ad=bc. ad=bc.

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.

How To

Given a rational equation, solve it.

  1. Factor all denominators in the equation.
  2. Find and exclude values that set each denominator equal to zero.
  3. Find the LCD.
  4. Multiply the whole equation by the LCD. If the LCD is correct, there will be no denominators left.
  5. Solve the remaining equation.
  6. Make sure to check solutions back in the original equations to avoid a solution producing zero in a denominator.

Example 4

Solving a Rational Equation without Factoring

Solve the following rational equation:

2 x 3 2 = 7 2x 2 x 3 2 = 7 2x
Try It #3

Solve the rational equation: 2 3x = 1 4 1 6x . 2 3x = 1 4 1 6x .

Example 5

Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: 1 x = 1 10 3 4x . 1 x = 1 10 3 4x .

Try It #4

Solve the rational equation: 5 2x + 3 4x = 7 4 . 5 2x + 3 4x = 7 4 .

Example 6

Solving Rational Equations with a Binomial in the Denominator

Solve the following rational equations and state the excluded values:

  1. 3 x6 = 5 x 3 x6 = 5 x
  2. x x3 = 5 x3 1 2 x x3 = 5 x3 1 2
  3. x x2 = 5 x2 1 2 x x2 = 5 x2 1 2
Try It #5

Solve 3 2x+1 = 4 3x+1 . 3 2x+1 = 4 3x+1 . State the excluded values.

Example 7

Solving a Rational Equation with Factored Denominators and Stating Excluded Values

Solve the rational equation after factoring the denominators: 2 x+1 1 x1 = 2x x 2 1 . 2 x+1 1 x1 = 2x x 2 1 . State the excluded values.

Try It #6

Solve the rational equation: 2 x2 + 1 x+1 = 1 x 2 x2 . 2 x2 + 1 x+1 = 1 x 2 x2 .

Finding a Linear Equation

Perhaps the most familiar form of a linear equation is the slope-intercept form, written as y=mx+b, y=mx+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.

m= y 2 y 1 x 2 x 1 m= y 2 y 1 x 2 x 1

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 .

Coordinate plane with the x and y axes ranging from negative 10 to 10.  Three linear functions are plotted: y = negative 3 times x minus 2; y = 2 times x plus 1; and y = x over 3 plus 2.
Figure 2

The Slope of a Line

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:

m= y 2 y 1 x 2 x 1 m= y 2 y 1 x 2 x 1

Example 8

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 ).

Analysis

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.

Try It #7

Find the slope of the line that passes through the points ( −2,6 ) ( −2,6 ) and ( 1,4 ). ( 1,4 ).

Example 9

Identifying the Slope and y-intercept of a Line Given an Equation

Identify the slope and y-intercept, given the equation y= 3 4 x4. y= 3 4 x4.

Analysis

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.

y y 1 =m( x x 1 ) y y 1 =m( x x 1 )

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.

The Point-Slope Formula

Given one point and the slope, the point-slope formula will lead to the equation of a line:

y y 1 =m( x x 1 ) y y 1 =m( x x 1 )

Example 10

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.

Analysis

Note that any point on the line can be used to find the equation. If done correctly, the same final equation will be obtained.

Try It #8

Given m=4, m=4, find the equation of the line in slope-intercept form passing through the point ( 2,5 ). ( 2,5 ).

Example 11

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.

Analysis

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.

y(3) = 7 3 (x0) y+3 = 7 3 x y = 7 3 x3 y(3) = 7 3 (x0) y+3 = 7 3 x y = 7 3 x3

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

Ax+By=C Ax+By=C

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.

Example 12

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.

Try It #9

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

x=c x=c

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.

m= 53 3( −3 ) = 2 0 m= 53 3( −3 ) = 2 0

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

y=c y=c

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.

m = −2(−2) 0(−2) = 0 2 = 0 m = −2(−2) 0(−2) = 0 2 = 0

Use any point for ( x 1 , y 1 ) ( x 1 , y 1 ) in the formula, or use the y-intercept.

y(−2) = 0(x3) y+2 = 0 y = −2 y(−2) = 0(x3) y+2 = 0 y = −2

The graph is a horizontal line through y=−2. y=−2. Notice that all of the y-coordinates are the same. See Figure 3.

Coordinate plane with the x-axis ranging from negative 7 to 4 and the y-axis ranging from negative 4 to 4.  The function y = negative 2 and the line x = negative 3 are plotted.
Figure 3 The line x = −3 is a vertical line. The line y = −2 is a horizontal line.

Example 13

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 ).

Try It #10

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.

Coordinate plane with the x-axis ranging from negative 8 to 8 in intervals of 2 and the y-axis ranging from negative 7 to 7.  Three functions are graphed on the same plot: y = 2 times x minus 3; y = 2 times x plus 1 and y = 2 times x plus 5.
Figure 4 Parallel lines

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 .

m1m2 = −1 3( 1 3 ) = −1 m1m2 = −1 3( 1 3 ) = −1
Coordinate plane with the x-axis ranging from negative 3 to 6 and the y-axis ranging from negative 2 to 5.  Two functions are graphed on the same plot: y = 3 times x minus 1 and y = negative x/3 minus 2.  Their intersection is marked by a box to show that it is a right angle.
Figure 5 Perpendicular lines

Example 14

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: 3y=4x+3 3y=4x+3and 3x4y=8. 3x4y=8.

Try It #11

Graph the two lines and determine whether they are parallel, perpendicular, or neither: 2yx=10 2yx=10 and 2y=x+4. 2y=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.

How To

Given an equation for a line, write the equation of a line parallel or perpendicular to it.

  1. Find the slope of the given line. The easiest way to do this is to write the equation in slope-intercept form.
  2. Use the slope and the given point with the point-slope formula.
  3. Simplify the line to slope-intercept form and compare the equation to the given line.

Example 15

Writing the Equation of a Line Parallel to a Given Line Passing Through a Given Point

Write the equation of line parallel to a 5x+3y=1 5x+3y=1 and passing through the point ( 3,5 ). ( 3,5 ).

Try It #12

Find the equation of the line parallel to 5x=7+y 5x=7+y and passing through the point ( −1,−2 ). ( −1,−2 ).

Example 16

Finding the Equation of a Line Perpendicular to a Given Line Passing Through a Given Point

Find the equation of the line perpendicular to 5x3y+4=05x3y+4=0 and passing through the point ( 4,1 ). (4,1).

2.2 Section Exercises

Verbal

1.

What does it mean when we say that two lines are parallel?

2.

What is the relationship between the slopes of perpendicular lines (assuming neither is horizontal nor vertical)?

3.

How do we recognize when an equation, for example y=4x+3, y=4x+3, will be a straight line (linear) when graphed?

4.

What does it mean when we say that a linear equation is inconsistent?

5.

When solving the following equation:

2 x5 = 4 x+1 2 x5 = 4 x+1

explain why we must exclude x=5 x=5 and x=−1 x=−1 as possible solutions from the solution set.

Algebraic

For the following exercises, solve the equation for x. x.

6.

7x+2=3x9 7x+2=3x9

7.

4x3=5 4x3=5

8.

3(x+2)12=5(x+1) 3(x+2)12=5(x+1)

9.

125(x+3)=2x5 125(x+3)=2x5

10.

1 2 1 3 x= 4 3 1 2 1 3 x= 4 3

11.

x 3 3 4 = 2x+3 12 x 3 3 4 = 2x+3 12

12.

2 3 x+ 1 2 = 31 6 2 3 x+ 1 2 = 31 6

13.

3(2x1)+x=5x+3 3(2x1)+x=5x+3

14.

2x 3 3 4 = x 6 + 21 4 2x 3 3 4 = x 6 + 21 4

15.

x+2 4 x1 3 =2 x+2 4 x1 3 =2

For the following exercises, solve each rational equation for x. x. State all x-values that are excluded from the solution set.

16.

3 x 1 3 = 1 6 3 x 1 3 = 1 6

17.

2 3 x+4 = x+2 x+4 2 3 x+4 = x+2 x+4

18.

3 x2 = 1 x1 + 7 (x1)(x2) 3 x2 = 1 x1 + 7 (x1)(x2)

19.

3x x1 +2= 3 x1 3x x1 +2= 3 x1

20.

5 x+1 + 1 x3 = 6 x 2 2x3 5 x+1 + 1 x3 = 6 x 2 2x3

21.

1 x = 1 5 + 3 2x 1 x = 1 5 + 3 2x

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.

22.

( 0,3 ) ( 0,3 )with a slope of 2 3 2 3

23.

( 1,2 ) ( 1,2 )with a slope of 4 5 4 5

24.

x-intercept is 1, and ( −2,6 ) ( −2,6 )

25.

y-intercept is 2, and ( 4,−1 ) ( 4,−1 )

26.

(−3,10) (−3,10)and (5,−6) (5,−6)

27.

( 1,3 )  and  ( 5,5 ) ( 1,3 )  and  ( 5,5 )

28.

parallel to y=2x+5 y=2x+5and passes through the point ( 4,3 ) ( 4,3 )

29.

perpendicular to 3y=x4 3y=x4and passes through the point ( −2,1 ) ( −2,1 ).

For the following exercises, find the equation of the line using the given information.

30.

( 2,0 ) ( 2,0 )and ( −2,5 ) ( −2,5 )

31.

( 1,7 ) ( 1,7 )and ( 3,7 ) ( 3,7 )

32.

The slope is undefined and it passes through the point ( 2,3 ). ( 2,3 ).

33.

The slope equals zero and it passes through the point ( 1,−4 ). ( 1,−4 ).

34.

The slope is 3 4 3 4 and it passes through the point (1,4). (1,4).

35.

( −1,3 ) ( −1,3 )and ( 4,−5 ) ( 4,−5 )

Graphical

For the following exercises, graph the pair of equations on the same axes, and state whether they are parallel, perpendicular, or neither.

36.

y=2x+7 y= 1 2 x4 y=2x+7 y= 1 2 x4

37.

3x2y=5 6y9x=6 3x2y=5 6y9x=6

38.

y= 3x+1 4 y=3x+2 y= 3x+1 4 y=3x+2

39.

x=4 y=−3 x=4 y=−3

Numeric

For the following exercises, find the slope of the line that passes through the given points.

40.

( 5,4 ) ( 5,4 )and ( 7,9 ) ( 7,9 )

41.

( −3,2 ) ( −3,2 )and ( 4,−7 ) ( 4,−7 )

42.

( −5,4 ) ( −5,4 )and ( 2,4 ) ( 2,4 )

43.

( −1,−2 ) ( −1,−2 )and ( 3,4 ) ( 3,4 )

44.

( 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.

45.

( −1,3 )  and  ( 5,1 ) ( −2,3 )  and  ( 0,9 ) ( −1,3 )  and  ( 5,1 ) ( −2,3 )  and  ( 0,9 )

46.

( 2,5 )  and  ( 5,9 ) ( −1,−1 )  and  ( 2,3 ) ( 2,5 )  and  ( 5,9 ) ( −1,−1 )  and  ( 2,3 )

Technology

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.

47.

0.537x2.19y=100 0.537x2.19y=100

48.

4,500x200y=9,528 4,500x200y=9,528

49.

20030y x =70 20030y x =70

Extensions

50.

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.

51.

Starting with the standard form of an equation Ax + By = C, Ax + By = C, solve this expression for y in terms of A,B,C, A,B,C,and x. x. Then put the expression in slope-intercept form.

52.

Use the above derived formula to put the following standard equation in slope intercept form: 7x5y=25. 7x5y=25.

53.

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)

54.

Find the slopes of the diagonals in the previous exercise. Are they perpendicular?

Real-World Applications

55.

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.

56.

If the profit equation for a small business selling x x number of item one and y y number of item two is p=3x+4y, p=3x+4y, 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+.25x, y=45+.25x, where x x is the number of miles traveled.

57.

What is your cost if you travel 50 mi?

58.

If your cost were $63.75, $63.75,how many miles were you charged for traveling?

59.

Suppose you have a maximum of $100 to spend for the car rental. What would be the maximum number of miles you could travel?

Citation/Attribution

Want to cite, share, or modify this book? This book is Creative Commons Attribution License 4.0 and you must attribute OpenStax.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/college-algebra/pages/1-introduction-to-prerequisites
Citation information

© Feb 10, 2020 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License 4.0 license. The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.