# 0.7: Review - Solving Linear Equations

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## Solving Basic Linear Equations

An equation is a statement indicating that two algebraic expressions are equal. A linear equation with one variable, $$x$$, is an equation that can be written in the standard form $$ax + b = 0$$ where $$a$$ and $$b$$ are real numbers and $$a ≠ 0$$. For example

$$3 x - 12 = 0$$

A solution to a linear equation is any value that can replace the variable to produce a true statement. The variable in the linear equation $$3x − 12 = 0$$ is $$x$$ and the solution is $$x = 4$$. To verify this, substitute the value $$4$$ in for $$x$$ and check that you obtain a true statement.

\begin{aligned} 3 x - 12 & = 0 \\ 3 ( \color{Cerulean}{4}\color{Black}{ )} - 12 & = 0 \\ 12 - 12 & = 0 \\ 0 & = 0 \:\: \color{Cerulean}{✓} \end{aligned}

Alternatively, when an equation is equal to a constant, we may verify a solution by substituting the value in for the variable and showing that the result is equal to that constant. In this sense, we say that solutions “satisfy the equation.”

Example $$\PageIndex{1}$$:

Is $$a=−\frac{1}{2}$$ a solution to $$−10a+5=25$$?

Solution

Recall that when evaluating expressions, it is a good practice to first replace all variables with parentheses, and then substitute the appropriate values. By making use of parentheses, we avoid some common errors when working the order of operations.

$$- 10 a + 5 = - 10 ( \color{Cerulean}{- \frac { 1 } { 2 }} \color{Black}{ ) +} 5 = 5 + 5 = 10 \neq 25\:\: \color{red}{✗}$$

No, $$a=−\frac{1}{2}$$ does not satisfy the equation.

Developing techniques for solving various algebraic equations is one of our main goals in algebra. This section reviews the basic techniques used for solving linear equations with one variable. We begin by defining equivalent equations as equations with the same solution set.

\left. \begin{aligned} 3 x - 5 & = 16 \\ 3 x & = 21 \\ x & = 7 \end{aligned} \right\} \quad \color{Cerulean}{Equivalent \:equations}

Here we can see that the three linear equations are equivalent because they share the same solution set, namely, $$\{7\}$$. To obtain equivalent equations, use the following properties of equality. Given algebraic expressions $$A$$ and $$B$$, where $$c$$ is a nonzero number:

Addition property of equality: If $$A=B$$, then $$A\color{Cerulean}{+c}\color{Black}{ = } B\color{Cerulean}{+c}$$ If $$A=B$$, then $$A\color{Cerulean}{-c}\color{Black}{ = }B\color{Cerulean}{-c}$$ If $$A=B$$, then $$\color{Cerulean}{c}\color{Black}{A} = \color{Cerulean}{c}\color{Black}{B}$$ If $$A=B$$, then $$\dfrac{A}{\color{Cerulean}{c}}\color{Black}{ = }\dfrac{B}{\color{Cerulean}{c}}$$

CAUTION: Multiplying or dividing both sides of an equation by $$0$$ should be carefully avoided. Dividing by $$0$$ is undefined and multiplying both sides by $$0$$ results in the equation $$0 = 0$$.

We solve algebraic equations by isolating the variable with a coefficient of 1. If given a linear equation of the form $$ax + b = c$$, then we can solve it in two steps. 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 the solution in the following examples is left to the reader.

Example $$\PageIndex{2}$$:

Solve: $$7x − 2 = 19$$.

Solution

\begin{aligned} 7 x - 2 & = 19 \\ 7 x - 2 \color{Cerulean}{+ 2} & = 19 \color{Cerulean}{+ 2} \quad Add\: 2\: to\: both\: sides. \\ 7 x & = 21 \\ \frac { 7 x } { \color{Cerulean}{7} } & = \frac { 21 } { \color{Cerulean}{7} } \quad \color{Cerulean}{Divide\: both\: sides\: by\: 7.} \\ x & = 3 \qquad \text{The solution set is: } \{ 3 \} \end{aligned}

Example $$\PageIndex{3}$$:

Solve: $$56 = 8 + 12y$$.

Solution

When no sign precedes the term, it is understood to be positive. In other words, think of this as $$56 = +8 + 12y$$. Therefore, we begin by subtracting $$8$$ on both sides of the equal sign.

\begin{aligned} 56 \color{Cerulean}{- 8} & = 8 + 12 y \color{Cerulean}{- 8} \\ 48 & = 12 y \\ \frac { 48 } { \color{Cerulean}{12} } & = \frac { 12 y } { \color{Cerulean}{12} } \\ 4 & = y \qquad \text{The solution set is: } \{ 4 \} \end{aligned}

It does not matter on which side we choose to isolate the variable because the symmetric property states that $$4 = y$$ is equivalent to $$y = 4$$.

Example $$\PageIndex{4}$$:

Solve: $$\frac { 5 } { 3 } x + 2 = - 8$$.

Solution

Isolate the variable term using the addition property of equality, and then multiply both sides of the equation by the reciprocal of the coefficient $$\frac{5}{3}$$ .

\begin{aligned} \frac { 5 } { 3 } x + 2 & = - 8 \\ \frac { 5 } { 3 } x + 2 \color{Cerulean}{- 2} & = - 8 \color{Cerulean}{- 2}\quad \color{Cerulean}{Subtract\: 2\: on\: both\: sides.} \\ \frac { 5 } { 3 } x & = - 10 \\ \color{Cerulean}{\frac { 3 } { 5 }} \color{Black}{ \cdot} \frac { 5 } { 3 } x & = \color{Cerulean}{\frac { 3 } { \cancel{5} }} \color{Black}{\cdot} ( \overset{-2}{\cancel{-10}} )\quad \color{Cerulean}{Multiply \:both \:sides\: by\: \frac{3}{5}.} \\ 1x & = 3 \cdot ( - 2 ) \\ x & = - 6 \qquad \text{The solution set is: } \{ -6 \} \end{aligned}

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation. Try It $$\PageIndex{4}$$

Solve: $$\dfrac { 2 } { 3 } x + \dfrac { 1 } { 2 } = - \dfrac { 5 } { 6 }$$.

$$x=-2$$

## General Guidelines for Solving Linear Equations How To: Solve a Linear Equation

Step 1: Simplify each side of the equation separately using the order of operations and combining all like terms on the same side of the equal sign.

Step 2: Use the appropriate properties of equality (add or subtract) to combine like terms on opposite sides of the equal sign. The goal is to obtain the variable term on one side of the equation and the constant term on the other.

Step 3: Divide or multiply as needed to isolate the variable.

Step 4: Check to see if the answer solves the original equation.

At this point in our study of algebra the use of the properties of equality should seem routine. Therefore, displaying these steps in this text, usually in blue, becomes optional.

Example $$\PageIndex{5}$$:

Solve: $$- 4 a + 2 - a = 1$$.

Solution

First combine the like terms on the left side of the equal sign.

\begin{aligned} - 4 a + 2 - a = 1 & \quad \color{Cerulean}{ Combine\: same-side\: like\: terms.} \\ - 5 a + 2 = 1 & \quad\color{Cerulean} { Subtract\: 2\: on\: both\: sides.} \\ - 5 a = - 1 & \quad\color{Cerulean} { Divide\: both\: sides\: by\: - 5.} \\ a = \frac { - 1 } { - 5 } = \frac { 1 } { 5 } & \qquad \text{The solution set is: } \{ \tfrac{1}{5} \} \end{aligned}

Always use the original equation to check to see if the solution is correct.

\begin{aligned} - 4 a + 2 - a & = - 4 \left( \color{OliveGreen}{\frac { 1 } { 5 }} \right) + 2 - \color{OliveGreen}{\frac { 1 } { 5 }} \\ & = - \frac { 4 } { 5 } + \frac { 2 } { 1 } \cdot \color{Cerulean}{\frac { 5 } { 5 }}\color{Black}{ -} \frac { 1 } { 5 } \\ & = \frac { - 4 + 10 + 1 } { 5 } \\ & = \frac { 5 } { 5 } = 1 \:\:\color{Cerulean}{✓} \end{aligned}

Given a linear equation in the form $$ax + b = cx + d$$, we begin the solving process by combining like terms on opposite sides of the equal sign. To do this, use the addition or subtraction property of equality to place like terms on the same side so that they can be combined. In the examples that remain, the check is left to the reader.

Example $$\PageIndex{6}$$:

Solve: $$−2y − 3 = 5y + 11$$.

Solution

Subtract $$5y$$ on both sides so that we can combine the terms involving y on the left side.

$$\begin{array} { c } { - 2 y - 3 \color{Cerulean}{- 5 y}\color{Black}{ =} 5 y + 11 \color{Cerulean}{- 5 y} } \\ { - 7 y - 3 = 11 } \end{array}$$

From here, solve using the techniques developed previously.

\begin{aligned} - 7 y - 3 & = 11 \quad\color{Cerulean}{Add\: 3\: to\: both\: sides.} \\ - 7 y & = 14 \\ y & = \frac { 14 } { - 7 } \quad\color{Cerulean}{Divide\: both\: sides\: by\: -7.} \\ y & = - 2 \qquad \text{The solution set is: } \{ -2 \} \end{aligned}

Solving will often require the application of the distributive property.

Example $$\PageIndex{7}$$:

Solve: $$- \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x )$$.

Solution

Simplify the linear expressions on either side of the equal sign first.

\begin{aligned} - \frac { 1 } { 2 } ( 10 x - 2 ) + 3 = 7 ( 1 - 2 x ) & \quad\color{Cerulean} { Distribute } \\ - 5 x + 1 + 3 = 7 - 14 x & \quad\color{Cerulean} { Combine\: same-side\: like\: terms. } \\ - 5 x + 4 = 7 - 14 x & \quad\color{Cerulean} { Combine\: opposite-side\: like\: terms. } \\ 9 x = 3 & \quad\color{Cerulean} { Solve. } \\ x = \frac { 3 } { 9 } = \frac { 1 } { 3 } & \qquad \text{The solution set is: } {\Large\{} \small{\frac{1}{3}} {\Large\}} \end{aligned}

Example $$\PageIndex{8}$$:

Solve: $$5(3−a)−2(5−2a)=3$$.

Solution

Begin by applying the distributive property.

\begin{aligned} 5 ( 3 - a ) - 2 ( 5 - 2 a ) & = 3 \\ 15 - 5 a - 10 + 4 a & = 3 \\ 5 - a & = 3 \\ - a & = - 2 \end{aligned}

Here we point out that $$−a$$ is equivalent to $$−1a$$; therefore, we choose to divide both sides of the equation by $$−1$$.

$$\begin{array} { c } { - a = - 2 } \\ { \dfrac { - 1 a } { \color{Cerulean}{- 1} }\color{Black}{ =} \dfrac { - 2 } { \color{Cerulean}{- 1} } } \\ { a = 2 } & \qquad \text{The solution set is: } \{ 2 \} \end{array}$$

Alternatively, we can multiply both sides of $$−a=−2$$ by negative one and achieve the same result.

\begin{aligned} - a & = - 2 \\ \color{Cerulean}{( - 1 )}\color{Black}{ (} - a ) & = \color{Cerulean}{( - 1 )}\color{Black}{ (} - 2 ) \\ a & = 2 \end{aligned} Try It $$\PageIndex{8}$$

Solve: $$6 - 3 ( 4 x - 1 ) = 4 x - 7$$.

$$x=1$$

There are three different types of equations. Up to this point, we have been solving conditional equations. These are equations that are true for particular values. An identity is an equation that is true for all possible values of the variable. For example,

$$x = x \quad\color{Cerulean}{Identity}$$

has a solution set consisting of all real numbers, $$ℝ$$. A contradiction is an equation that is never true and thus has no solutions. For example,

$$x+1=x\quad\color{Cerulean}{Contradiction}$$

has no solution. We use the empty set, $$Ø$$, to indicate that there are no solutions.

If the end result of solving an equation is a true statement, like $$0 = 0$$, then the equation is an identity and any real number is a solution. If solving results in a false statement, like $$0 = 1$$, then the equation is a contradiction and there is no solution.

Example $$\PageIndex{9}$$:

Solve: $$4 (x + 5) + 6 = 2 (2x + 3)$$.

Solution

\begin{aligned} 4 ( x + 5 ) + 6 & = 2 ( 2 x + 3 ) \\ 4 x + 20 + 6 & = 4 x + 6 \\ 4 x + 26 & = 4 x + 6 \\ 26 & = 6\:\: \color{red}{✗} \end{aligned}

Solving leads to a false statement; therefore, the equation is a contradiction and there is no solution.

Answer: $$Ø$$ or $$\{ \;\; \}$$

Example $$\PageIndex{10}$$:

Solve: $$3 (3y + 5) + 5 = 10 (y + 2) − y$$.

Solution

\begin{aligned} 3 ( 3 y + 5 ) + 5 & = 10 ( y + 2 ) - y \\ 9 y + 15 + 5 & = 10 y + 20 - y \\ 9 y + 20 & = 9 y + 20 \\ 9 y & = 9 y \\ 0 & = 0 \:\:\color{Cerulean}{✓} \end{aligned}

Solving leads to a true statement; therefore, the equation is an identity and any real number is a solution.

Answer: $$ℝ$$

The coefficients of linear equations may be any real number, even decimals and fractions. When this is the case it is possible to use the multiplication property of equality to clear the fractional coefficients and obtain integer coefficients in a single step. If given fractional coefficients, then multiply both sides of the equation by the least common multiple of the denominators (LCD).

Example $$\PageIndex{11}$$:

Solve: $$\dfrac { 1 } { 3 } x + \dfrac { 1 } { 5 } = \dfrac { 1 } { 5 } x - 1$$.

Solution

Clear the fractions by multiplying both sides by the least common multiple of the given denominators. In this case, it is the $$LCD (3, 5) = 15$$.

\begin{aligned} \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 3 } x + \frac { 1 } { 5 } \right) & = \color{Cerulean}{15}\color{Black}{ \cdot} \left( \frac { 1 } { 5 } x - 1 \right) \quad \color{Cerulean}{Multiply\: both\: sides\: by\: 15.} \\ \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 3 } x + \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } & = \color{Cerulean}{15}\color{Black}{ \cdot} \frac { 1 } { 5 } x - \color{Cerulean}{15}\color{Black}{ \cdot} 1\quad\color{Cerulean}{Simplify.} \\ 5 x + 3 & = 3 x - 15\quad\quad\quad\color{Cerulean}{Solve.} \\ 2 x & = - 18 \\ x & = \frac { - 18 } { 2 } = - 9 \qquad \text{The solution set is: } \{ -9 \} \end{aligned}

It is important to know that this technique only works for equations. Do not try to clear fractions when simplifying expressions. As a reminder:

Expression Equation
$$\dfrac { 1 } { 2 } x + \dfrac { 5 } { 3 }$$ $$\dfrac { 1 } { 2 } x + \dfrac { 5 } { 3 }=0$$

We simplify expressions and solve equations. If you multiply an expression by $$6$$, you will change the problem. However, if you multiply both sides of an equation by $$6$$, you obtain an equivalent equation.

Incorrect Correct

$$\dfrac { 1 } { 2 } x + \dfrac { 5 } { 3 }$$

\begin{aligned} \neq & \color{red}{6 \cdot}\color{Black}{ \left( \frac { 1 } { 2 } x + \frac { 5 } { 3 } \right)} \\ = & 3 x + 10 \quad \color{red}{✗} \end{aligned}

\begin{aligned} \dfrac { 1 } { 2 } x + \dfrac { 5 } { 3 } & = 0 \\ \color{Cerulean}{6 \cdot}\color{Black}{ \left( \frac { 1 } { 2 } x + \frac { 5 } { 3 } \right)} & = \color{Cerulean}{6 \cdot}\color{Black}{ 0} \\ 3 x + 10 & = 0\quad\color{Cerulean}{✓} \end{aligned}

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