# 2.2 Linear equations in one variable

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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 [link] .

## 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 $\text{\hspace{0.17em}}ax+b=0\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}5x+2=3x-6,$ we have the following:

$\begin{array}{ccc}\hfill 5x+2& =& 3x-6\hfill \\ \hfill 2x& =& -8\hfill \\ \hfill x& =& -4\hfill \end{array}$

The solution set consists of one number: $\text{\hspace{0.17em}}\left\{-4\right\}.\text{\hspace{0.17em}}$ 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 $\text{\hspace{0.17em}}5x-15=5\left(x-4\right),$ we have the following:

Indeed, $\text{\hspace{0.17em}}-15\ne \text{\hspace{0.17em}}-20.\text{\hspace{0.17em}}$ 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, $\text{\hspace{0.17em}}a\ne 0.$

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 $\text{\hspace{0.17em}}x=_________,$ 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: $\text{\hspace{0.17em}}a\left(b+c\right)=ab+ac.$
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.

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