Linear equations in one variable

Introduction

This chapter is about linear equations in one variable. We first review constants and variables, then learn what a linear equation in one variable is, what it means for a value to be a solution, and how to solve an equation using the properties of equality. We finish with equations that have no solution or infinitely many solutions.

Constants and variables

In algebra, we work with two kinds of quantities: constants and variables.

A constant is a fixed number, such as $8$, $-5$, or $2$.

A variable is a letter that stands for a number, such as $x$, $y$, or $u$. A variable can take different values; for example, $x$ might be $7$, $-6$, or $0$.

Linear equation in one variable

An equation that contains only one variable, and in which the exponent (power) of the variable is $1$, is called a linear equation in one variable.

For example, the following are linear equations in one variable:

$2x+5=3$,   $\dfrac{7}{3}y=-6$,   $\dfrac{6}{11}w=-6+2w$

In each one there is a single variable whose exponent is $1$.

The equation $\dfrac{7}{3}y^2=-6$ is not a linear equation, because the exponent of the variable is $2$.

Solution of an equation

The solution of an equation is the value of the variable that makes the equation true.

Example: For the equation $4x+2=14$, determine whether $x=3$ is a solution and whether $x=1$ is a solution.

Solution: Substitute each value into the equation and check whether the two sides are equal.

Substitute $x=3$:

$\begin{align*}4(3)+2&=14\\12+2&=14\\14&=14\end{align*}$

This statement is true, so $x=3$ is a solution of the equation.

Substitute $x=1$:

$\begin{align*}4(1)+2&=14\\4+2&=14\\6&=14\end{align*}$

This statement is not true, so $x=1$ is not a solution of the equation.

Solving an equation

Solving an equation means finding its solution. To solve an equation, we isolate the variable on one side of the equation using the properties of equality.

Properties of equality

Whatever you do to one side of an equation, you must do to the other side, so that the two sides stay equal.

Addition property: You can add the same number to both sides of an equation. If $a=b$, then $a+c=b+c$.

Subtraction property: You can subtract the same number from both sides of an equation. If $a=b$, then $a-c=b-c$.

Multiplication property: You can multiply both sides of an equation by the same number. If $a=b$, then $ac=bc$.

Division property: You can divide both sides of an equation by the same nonzero number. If $a=b$ and $c\ne 0$, then $\dfrac{a}{c}=\dfrac{b}{c}$. Never divide by $0$, since division by zero is undefined.

Solving linear equations

The following examples show how to solve linear equations by isolating the variable.

Example 1: Solve $7+2x=15$.

Solution: Subtract $7$ from both sides:

$\begin{align*}7+2x-7&=15-7\\2x&=8\end{align*}$

Divide both sides by $2$:

$\begin{align*}\dfrac{2x}{2}&=\dfrac{8}{2}\\x&=4\end{align*}$

Example 2: Solve $\dfrac{4}{7}u=8$.

Solution: Multiply both sides by $7$:

$\begin{align*}\dfrac{4}{7}u\cdot 7&=8\cdot 7\\4u&=56\end{align*}$

Divide both sides by $4$:

$\begin{align*}\dfrac{4u}{4}&=\dfrac{56}{4}\\u&=14\end{align*}$

Example 3: Solve $5y+42=-3(y-6)$.

Solution: Use the distributive property to remove the parentheses:

$5y+42=-3y+18$

Add $3y$ to both sides:

$\begin{align*}5y+3y+42&=18\\8y+42&=18\end{align*}$

Subtract $42$ from both sides:

$\begin{align*}8y&=18-42\\8y&=-24\end{align*}$

Divide both sides by $8$:

$y=-3$

Example 4: Solve $4v-2(v-7)=-4(-2v+6)+5$.

Solution: Use the distributive property to remove the parentheses on both sides:

$\begin{align*}4v-2v+14&=8v-24+5\\2v+14&=8v-19\end{align*}$

Subtract $8v$ from both sides and subtract $14$ from both sides:

$\begin{align*}2v-8v&=-19-14\\-6v&=-33\end{align*}$

Divide both sides by $-6$:

$v=\dfrac{-33}{-6}=\dfrac{11}{2}$

Example 5: Solve $-\dfrac{5}{3}=-\dfrac{4}{9}y+\dfrac{5}{6}$.

Solution: The equation has fractions. The denominators are $3$, $9$, and $6$, so the least common denominator (LCD) is $18$. Multiply every term by $18$ to clear the fractions:

$\begin{align*}18\cdot\left(-\dfrac{5}{3}\right)&=18\cdot\left(-\dfrac{4}{9}y\right)+18\cdot\dfrac{5}{6}\\-30&=-8y+15\end{align*}$

Subtract $15$ from both sides:

$\begin{align*}-30-15&=-8y\\-45&=-8y\end{align*}$

Divide both sides by $-8$:

$y=\dfrac{-45}{-8}=\dfrac{45}{8}$

Example 6: Solve $\dfrac{5u}{2}-\dfrac{13}{5}=\dfrac{9u}{5}-2$.

Solution: The denominators are $2$ and $5$, so the LCD is $10$. Multiply every term by $10$ to clear the fractions:

$\begin{align*}10\cdot\dfrac{5u}{2}-10\cdot\dfrac{13}{5}&=10\cdot\dfrac{9u}{5}-10\cdot 2\\25u-26&=18u-20\end{align*}$

Subtract $18u$ from both sides:

$\begin{align*}25u-18u-26&=-20\\7u-26&=-20\end{align*}$

Add $26$ to both sides:

$7u=6$

Divide both sides by $7$:

$u=\dfrac{6}{7}$

Example 7: Solve $\dfrac{7u}{3}-\dfrac{u}{6}=\dfrac{5u}{18}$.

Solution: Every term has the variable. The denominators are $3$, $6$, and $18$, so the LCD is $18$. Multiply every term by $18$ to clear the fractions:

$\begin{align*}18\cdot\dfrac{7u}{3}-18\cdot\dfrac{u}{6}&=18\cdot\dfrac{5u}{18}\\42u-3u&=5u\\39u&=5u\end{align*}$

Subtract $5u$ from both sides:

$34u=0$

Divide both sides by $34$:

$u=0$

Example 8: Solve $\dfrac{7v}{6}-8=\dfrac{13v}{10}$.

Solution: The denominators are $6$ and $10$, so the LCD is $30$. Multiply every term by $30$ to clear the fractions:

$\begin{align*}30\cdot\dfrac{7v}{6}-30\cdot 8&=30\cdot\dfrac{13v}{10}\\35v-240&=39v\end{align*}$

Subtract $39v$ from both sides:

$\begin{align*}35v-39v-240&=0\\-4v-240&=0\end{align*}$

Add $240$ to both sides, then divide by $-4$:

$\begin{align*}-4v&=240\\v&=-60\end{align*}$

Equations with no solution or infinitely many solutions

Not every equation has exactly one solution. Some equations have no solution, and some have infinitely many solutions.

When you solve such an equation, the variable cancels out. If you are left with a false statement (such as $0=-5$), the equation has no solution. If you are left with a true statement (such as $0=0$), the equation has infinitely many solutions.

Example 1: Solve $-6+4(x+2)=3(x-1)+x$.

Solution: Use the distributive property and simplify each side:

$\begin{align*}-6+4x+8&=3x-3+x\\4x+2&=4x-3\end{align*}$

Subtract $4x$ from both sides:

$2=-3$

This statement is not true, so the equation has no solution.

Example 2: Solve $-5(2y-4)=4(5-2y)-2y$.

Solution: Use the distributive property and simplify each side:

$\begin{align*}-10y+20&=20-8y-2y\\-10y+20&=20-10y\end{align*}$

Add $10y$ to both sides:

$20=20$

This statement is always true, so every value of $y$ is a solution; the equation has infinitely many solutions.