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3 2.2 Linear Equations in One Variable – College Algebra 2e | OpenStax

Linear Equations in One Variable

CA0522

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 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

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 xx 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=3x-6, we have the following:

5x+2=3x-6

2x=-8

x=-4

The solution set consists of one number: { −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 5x-15=5(x-4), we have the following:

5x-15=5x-20

5x-5x-15=5x-5x-20

-15\neq{-20}        False statement

Indeed, −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,  where a and b are real numbers, a≠0.

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=_________, 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.
  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 I:  Solving an Equation in One Variable

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

Solution

This equation can be written in the form ax+b=0 by subtracting 1919 from both sides. However, we may proceed to solve the equation in its original form by performing algebraic operations.

2x+7=19

2x=19-7

2x=12

x=\frac{12}{2}=6.

The solution is 6.

TRY IT #1

Solve the linear equation in one variable: 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).


Solution

This problem requires the distributive property to be applied twice, and then the properties of algebra are used to reach the final line, x=-\frac{5}{3}.

TRY IT #2
Solve the equation in one variable: -2(3x-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 23 or 72. A rational expression is the ratio, or quotient, of two polynomials. Here are three examples.

\frac{(x+1)}{(x^2-4)}, \frac{1}{(x-3)}, or \frac{4}{(x^2 + x -2)}

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: \frac{7}{2x}-\frac{5}{3x}=\frac{22}{3}.

Solution

We have three denominators; 2x,3x, and 3. The LCD must contain 2x,3x, and 3. An LCD of 6x 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 6x.

{6x}\frac{7}{2x}-{6x}\frac{5}{3x}={6x}\frac{22}{2}.

3(7)-2(5)=2(22x).  Use the distributive property and cancel out the common factors.

21-10=44x  Multiply remaining factors by each numerator.

11 = 44x

x=\frac{11}{44}=1/4.

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). 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−1, and 3x−3. First, factor all denominators. We then have x, (x−1), and 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). The xx in the first denominator is separate from the xx in the (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, one factor of (x−1), and the 3. Thus, the LCD is the following:

x(x-1)3=3x(x-1).

So, both sides of the equation would be multiplied by 3x(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 and x^2+2xOnce the second denominator is factored as x^2+2x there is a common factor of x in both denominators and the LCD is 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.

ab=cd

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) and b(c), which results in 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:

\frac{2}{x}-\frac{3}{2}=\frac{7}{(2x)}.

Solution
TRY IT #3

Solve the rational equation: \frac{2}{(3x0}=\frac{1}{4}-\frac{1}{(6x)}.

EXAMPLE 5:   Solving a Rational Equation by Factoring the Denominator

Solve the following rational equation: \frac{1}{x}=\frac{1}[10}-\frac{3}{(4x)}.

Solution

First find the common denominator. The three denominators in factored form are x,10=2⋅5, and 4x=2⋅2⋅x. The smallest expression that is divisible by each one of the denominators is 20x. Only x=0 is an excluded value. Multiply the whole equation by 20x.

{20x}\frac{1}{x}={20x}\frac{1}{10}-{20x}\frac{3}{4x}.
20=2x-15.   Simplify
20+15=2x
2x=35 => x = \frac{35}{2}.

The solution is \frac{35}{2}.

 

 

 

End of section 2.2

 

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