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1.7: Solving Linear Equations

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Skills to Develop

• Use the properties of equality to solve basic linear equations.
• Identify and solve conditional linear equations, identities, and contradictions.
• Clear fractions from equations.
• Set up and solve linear equations.

Solving Basic Linear Equations

An equation129 is a statement indicating that two algebraic expressions are equal. A linear equation with one variable130, $$x$$, 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 solution131 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 equations132 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 equality133. 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 $$\frac{A}{\color{Cerulean}{c}}\color{Black}{=}\frac{B}{\color{Cerulean}{c}}$$

Table 1.7.1

Note

Multiplying or dividing both sides of an equation by $$0$$ is 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 \end{aligned}

The solution is $$3$$.

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 \end{aligned}

It does not matter on which side we choose to isolate the variable because the symmetric property134 states that $$4 = y$$ is equivalent to $$y = 4$$.

The solution is $$4$$.

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 \end{aligned}

The solution is $$−6$$.

In summary, to retain equivalent equations, we must perform the same operation on both sides of the equation.

Exercise $$\PageIndex{1}$$

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

$$x=-2$$

General Guidelines for Solving Linear Equations

Typically linear equations are not given in standard form, and so solving them requires additional steps. When solving linear equations, the goal is to determine what value, if any, will produce a true statement when substituted in the original equation. Do this by isolating the variable using the following steps:

• Step 1: Simplify both sides of the equation using the order of operations and combine all like terms on the same side of the equal sign.
• Step 2: Use the appropriate properties of equality 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.

We will often encounter linear equations where the expressions on each side of the equal sign can be simplified. If this is the case, then it is best to simplify each side first before solving. Normally this involves combining same-side like terms.

Note

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 } \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}

The solution is $$\frac{1}{5}$$ .

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 \end{aligned}

The solution is $$−2$$.

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 } \end{aligned}

The solution is $$\frac{1}{3}$$ .

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 $$−1$$.

$$\begin{array} { c } { - a = - 2 } \\ { \frac { - 1 a } { \color{Cerulean}{- 1} }\color{Black}{ =} \frac { - 2 } { \color{Cerulean}{- 1} } } \\ { a = 2 } \end{array}$$

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}

The solution is $$2$$.

Exercise $$\PageIndex{2}$$

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 equations135. These are equations that are true for particular values. An identity136 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 contradiction137 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.

$$Ø$$

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.

$$ℝ$$

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

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 \end{aligned}

The solution is $$−9$$.

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
$$\frac { 1 } { 2 } x + \frac { 5 } { 3 }$$ $$\frac { 1 } { 2 } x + \frac { 5 } { 3 }=0$$

Table 1.7.2

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

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

\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} \frac { 1 } { 2 } x + \frac { 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}

Table 1.7.3

Applications Involving Linear Equations

Algebra simplifies the process of solving real-world problems. This is done by using letters to represent unknowns, restating problems in the form of equations, and by offering systematic techniques for solving those equations. To solve problems using algebra, first translate the wording of the problem into mathematical statements that describe the relationships between the given information and the unknowns. Usually, this translation to mathematical statements is the difficult step in the process. The key to the translation is to carefully read the problem and identify certain key words and phrases.

Key Words Translation
Sum, increased by, more than, plus, added to, total $$+$$
Difference, decreased by, subtracted from, less, minus $$-$$
Product, multiplied by, of times, twice $$\cdot$$
Quotient, divided by, ratio, per $$÷$$
Is, total, result $$=$$

Table 1.7.4

When translating sentences into mathematical statements, be sure to read the sentence several times and parse out the key words and phrases. It is important to first identify the variable, “let x represent…” and state in words what the unknown quantity is. This step not only makes our work more readable, but also forces us to think about what we are looking for.

Example $$\PageIndex{12}$$:

When $$6$$ is subtracted from twice the sum of a number and $$8$$ the result is $$5$$. Find the number.

Solution:

Let n represent the unknown number. Figure 1.7.1

To understand why we included the parentheses in the set up, you must study the structure of the following two sentences and their translations:

 “twice the sum of a number and 8” 2(n+8) “the sum of twice a number and 8” 2n+8

Table 1.7.5

The key was to focus on the phrase “twice the sum,” this prompted us to group the sum within parentheses and then multiply by $$2$$. After translating the sentence into a mathematical statement we then solve.

\begin{aligned} 2 ( n + 8 ) - 6 & = 5 \\ 2 n + 16 - 6 & = 5 \\ 2 n + 10 & = 5 \\ 2 n & = - 5 \\ n & = \frac { - 5 } { 2 } \end{aligned}

Check.

\begin{aligned} 2 ( n + 8 ) - 6 & = 2 \left( \color{Cerulean}{- \frac { 5 } { 2 }}\color{Black}{ +} 8 \right) - 6 \\ & = 2 \left( \frac { 11 } { 2 } \right) - 6 \\ & = 11 - 6 \\ & = 5 \quad\color{Cerulean}{✓}\end{aligned}

The number is $$−\frac{5}{2}$$.

General guidelines for setting up and solving word problems follow.

• Step 1: Read the problem several times, identify the key words and phrases, and organize the given information.
• Step 2: Identify the variables by assigning a letter or expression to the unknown quantities.
• Step 3: Translate and set up an algebraic equation that models the problem.
• Step 4: Solve the resulting algebraic equation.
• Step 5: Finally, answer the question in sentence form and make sure it makes sense (check it).

For now, set up all of your equations using only one variable. Avoid two variables by looking for a relationship between the unknowns.

Example $$\PageIndex{13}$$:

A rectangle has a perimeter measuring $$92$$ meters. The length is $$2$$ meters less than $$3$$ times the width. Find the dimensions of the rectangle.

Solution:

The sentence “The length is 2 meters less than 3 times the width,” gives us the relationship between the two variables.

Let $$w$$ represent the width of the rectangle.

Let $$3w−2$$ represent the length. Figure 1.7.2

The sentence “A rectangle has a perimeter measuring 92 meters” suggests an algebraic set up. Substitute $$92$$ for the perimeter and the expression $$3w−2$$ for the length into the appropriate formula as follows:

$$P = \quad2 l\:\:\:\:\: +\:\:\: 2 w$$

$$\color{Cerulean}{\downarrow} \quad\:\:\:\quad \color{Cerulean}{\downarrow}\quad\quad\quad\quad$$

$$\color{OliveGreen}{92}\color{Black}{ =} 2 (\color{OliveGreen}{ 3 w - 2}\color{Black}{ )} + 2 w$$

Once you have set up an algebraic equation with one variable, solve for the width, $$w$$.

$$\begin{array} { l } { 92 = 2 ( 3 w - 2 ) + 2 w \color{Cerulean}{ Distribute. } } \\ { 92 = 6 w - 4 + 2 w \quad\: \color{Cerulean} { Combine\: like\: terms. } } \\ { 92 = 8 w - 4 \quad\quad\quad\:\:\:\color{Cerulean} { Solve\: for\: w. } } \\ { 96 = 8 w } \\ { 12 = w } \end{array}$$

Use $$3w−2$$ to find the length.

$$l = 3 w - 2 = 3 ( \color{OliveGreen}{12}\color{Black}{ )} - 2 = 36 - 2 = 34$$

To check, make sure the perimeter is $$92$$ meters.

\begin{aligned} P & = 2 l + 2 w \\ & = 2 ( 34 ) + 2 ( 12 ) \\ & = 68 + 24 \\ & = 92 \end{aligned}

The rectangle measures $$12$$ meters by $$34$$ meters.

Example $$\PageIndex{14}$$:

Given a $$4 \frac{3}{8}$$% annual interest rate, how long will it take $$2,500$$ to yield $$437.50$$ in simple interest?

Solution:

Let t represent the time needed to earn $$437.50$$ at $$4 \frac{3}{8}$$%. Organize the information needed to use the formula for simple interest, $$I=prt$$.

 Given interest for the time period: $$I=437.50$$ Given principal: $$p=2,500$$ Given rate: $$r= 4 \frac{3}{8}$$% $$=4.375$$ %$$=0.04375$$

Table 1.7.6

Next, substitute all of the known quantities into the formula and then solve for the only unknown, t.

\begin{aligned} I & = p r t \\ \color{OliveGreen}{437.50} & \color{Black}{=} \color{OliveGreen}{2500}\color{Black}{ (}\color{OliveGreen}{ 0.04375}\color{Black}{ )} t \\ 437.50 & = 109.375 t \\ \frac{437.50}{\color{Cerulean}{109.375}} & \color{Black}{=} \frac { 109.375 t } { \color{Cerulean}{109.375} } \\ 4 & = t \end{aligned}

It takes $$4$$ years for $$2,500$$ invested at $$4 \frac{3}{8}$$% to earn $$437.50$$ in simple interest.

Example $$\PageIndex{15}$$:

Susan invested her total savings of $$12,500$$ in two accounts earning simple interest. Her mutual fund account earned $$7$$% last year and her CD earned $$4.5$$%. If her total interest for the year was $$670$$, how much was in each account?

Solution:

The relationship between the two unknowns is that they total $$12,500$$. When a total is involved, a common technique used to avoid two variables is to represent the second unknown as the difference of the total and the first unknown.

Let $$x$$ represent the amount invested in the mutual fund.

Let $$12,500 − x$$ represent the remaining amount invested in the CD.

Organize the data.

 Interest earned in the mutual fund: \begin{aligned} I & = p r t \\ & = x \cdot 0.07 \cdot 1 \\ & = \color{OliveGreen}{0.07 x} \end{aligned} Interest earned in the CD: \begin{aligned} I & = p r t \\ & = ( 12,500 - x ) \cdot 0.045 \cdot 1 \\ & = \color{OliveGreen}{0.045 ( 12,500 - x )} \end{aligned} Total interest: $$\color{OliveGreen}{670}$$

Table 1.7.7

The total interest is the sum of the interest earned from each account.

$$\color{Cerulean}{mutual fund interest + CD interest = total interest}$$

$$0.07x + 0.045(12,500−x) = 670$$

This equation models the problem with one variable. Solve for $$x$$.

\begin{aligned} 0.07 x + 0.045 ( 12,500 - x ) & = 670 \\ 0.07 x + 562.5 - 0.045 x & = 670 \\ 0.025 x + 562.5 & = 670 \\ 0.025 x & = 107.5 \\ x & = \frac { 107.5 } { 0.025 } \\ x & = 4,300 \end{aligned}

Use $$12,500−x$$ to find the amount in the CD.

$$12,500−x=12,500−\color{OliveGreen}{4,300}\color{Black}{=}8,200$$

Susan invested $$4,300$$ at $$7$$% in a mutual fund and $$8,200$$ at $$4.5$$% in a CD.

Key Takeaways

• Solving general linear equations involves isolating the variable, with coefficient $$1$$, on one side of the equal sign. To do this, first use the appropriate equality property of addition or subtraction to isolate the variable term on one side of the equal sign. Next, isolate the variable using the equality property of multiplication or division. Finally, check to verify that your solution solves the original equation.
• If solving a linear equation leads to a true statement like $$0 = 0$$, then the equation is an identity and the solution set consists of all real numbers, $$ℝ$$.
• If solving a linear equation leads to a false statement like $$0 = 5$$, then the equation is a contradiction and there is no solution, $$Ø$$.
• Clear fractions by multiplying both sides of an equation by the least common multiple of all the denominators. Distribute and multiply all terms by the LCD to obtain an equivalent equation with integer coefficients.
• Simplify the process of solving real-world problems by creating mathematical models that describe the relationship between unknowns. Use algebra to solve the resulting equations.

Exercise $$\PageIndex{3}$$

Determine whether or not the given value is a solution.

1. $$−5x + 4 = −1 ; x = −1$$
2. $$4x − 3 = −7 ; x = −1$$
3. $$3y − 4 = 5; y = \frac{9}{3}$$
4. $$−2y + 7 = 12 ; y = −\frac{5}{2}$$
5. $$3a − 6 = 18 − a; a = −3$$
6. $$5 (2t − 1) = 2 − t; t = 2$$
7. $$ax − b = 0; x = \frac{b}{a}$$
8. $$ax + b = 2b; x = \frac{b}{a}$$

1. No

3. Yes

5. No

7. Yes

Exercise $$\PageIndex{4}$$

Solve.

1. $$5x − 3 = 27$$
2. $$6x − 7 = 47$$
3. $$4x + 13 = 35$$
4. $$6x − 9 = 18$$
5. $$9a + 10 = 10$$
6. $$5 − 3a = 5$$
7. $$−8t + 5 = 15$$
8. $$−9t + 12 = 33$$
9. $$\frac{2}{3} x + \frac{1}{2} = 1$$
10. $$\frac{3}{8} x + \frac{5}{4} = \frac{3}{2}$$
11. $$\frac{1 − 3y}{5} = 2$$
12. $$\frac{2 − 5y}{6} = −8$$
13. $$7 − y = 22$$
14. $$6 − y = 12$$
15. Solve for $$x: ax − b = c$$
16. Solve for $$x: ax + b = 0$$

1. $$6$$

3. $$\frac{11}{2}$$

5. $$0$$

7. $$−\frac{5}{4}$$

9. $$\frac{3}{4}$$

11. $$−3$$

13. $$−15$$

15. $$x = \frac{b+c}{a}$$

Exercise $$\PageIndex{5}$$

Solve.

1. $$6x − 5 + 2x = 19$$
2. $$7 − 2x + 9 = 24$$
3. $$12x − 2 − 9x = 5x + 8$$
4. $$16 − 3x − 22 = 8 − 4x$$
5. $$5y − 6 − 9y = 3 − 2y + 8$$
6. $$7 − 9y + 12 = 3y + 11 − 11y$$
7. $$3 + 3a − 11 = 5a − 8 − 2a$$
8. $$2 − 3a = 5a + 7 − 8a$$
9. $$\frac{1}{3} x −\frac{3}{2} + \frac{5}{2} x = \frac{5}{6} x + \frac{1}{4}$$
10. $$\frac{5}{8} + \frac{1}{5} x −\frac{3}{4} = \frac{3}{10} x − \frac{1}{4}$$
11. $$1.2x − 0.5 − 2.6x = 2 − 2.4x$$
12. $$1.59 − 3.87x = 3.48 − 4.1x − 0.51$$
13. $$5 − 10x = 2x + 8 − 12x$$
14. $$8x − 3 − 3x = 5x − 3$$
15. $$5 (y + 2) = 3 (2y − 1) + 10$$
16. $$7 (y − 3) = 4 (2y + 1) − 21$$
17. $$7 − 5 (3t − 9) = 22$$
18. $$10 − 5 (3t + 7) = 20$$
19. $$5 − 2x = 4 − 2 (x − 4)$$
20. $$2 (4x − 5) + 7x = 5 (3x − 2)$$
21. $$4 (4a − 1) = 5 (a − 3) + 2 (a − 2)$$
22. $$6 (2b − 1) + 24b = 8 (3b − 1)$$
23. $$\frac{2}{3} (x + 18) + 2 = \frac{1}{3} x − 13$$
24. $$\frac{2}{5} x − \frac{1}{2} (6x − 3) = \frac{4}{3}$$
25. $$1.2 (2x + 1) + 0.6x = 4x$$
26. $$6 + 0.5 (7x − 5) = 2.5x + 0.3$$
27. $$5 (y + 3) = 15 (y + 1) − 10y$$
28. $$3 (4 − y) − 2 (y + 7) = −5y$$
29. $$\frac{1}{5} (2a + 3) −\frac{1}{2} = \frac{1}{3} a + \frac{1}{10}$$
30. $$\frac{3}{2} a = \frac{3}{4} (1 + 2a) −\frac{1}{5} (a + 5)$$
31. $$6 − 3 (7x + 1) = 7 (4 − 3x)$$
32. $$6 (x − 6) − 3 (2x − 9) = −9$$
33. $$\frac{3}{4} (y − 2) + \frac{2}{3} (2y + 3) = 3$$
34. $$\frac{5}{4} − \frac{1}{2} (4y − 3) = \frac{2}{5} (y − 1)$$
35. $$−2 (3x + 1) − (x − 3) = −7x + 1$$
36. $$6 (2x + 1) − (10x + 9) = 0$$
37. Solve for $$w: P = 2l + 2w$$
38. Solve for $$a: P = a + b + c$$
39. Solve for $$t: D = rt$$
40. Solve for $$w: V = lwh$$
41. Solve for $$b: A = \frac{1}{2} bh$$
42. Solve for $$a:s = \frac{1}{2}at^{2}$$
43. Solve for $$a: A = \frac{1}{2}h (a + b)$$
44. Solve for $$h: V = \frac{1}{3}πr^{2}h$$
45. Solve for $$F: C = \frac{5}{9} (F − 32)$$
46. Solve for $$x: ax + b = c$$

1. $$3$$

3. $$−5$$

5. $$−\frac{17}{2}$$

7. $$ℝ$$

9. $$\frac{7}{8}$$

11. $$2.5$$

13. $$Ø$$

15. $$3$$

17. $$2$$

19. $$Ø$$

21. $$−\frac{5}{3}$$

23. $$−81$$

25. $$1.2$$

27. $$ℝ$$

29. $$0$$

31. $$Ø$$

33. $$\frac{6}{5}$$

35. $$ℝ$$

37. $$w = \frac{P − 2l}{2}$$

39. $$t = \frac{D}{r}$$

41. $$b = \frac{2A}{h}$$

43. $$a = \frac{2A}{h} − b$$

45. $$F = \frac{9}{5} C + 32$$

Exercise $$\PageIndex{6}$$

Set up an algebraic equation then solve.

Number Problems

1. When $$3$$ is subtracted from the sum of a number and $$10$$ the result is $$2$$. Find the number.
2. The sum of $$3$$ times a number and $$12$$ is equal to $$3$$. Find the number.
3. Three times the sum of a number and $$6$$ is equal to $$5$$ times the number. Find the number.
4. Twice the sum of a number and $$4$$ is equal to $$3$$ times the sum of the number and $$1$$. Find the number.
5. A larger integer is $$1$$ more than $$3$$ times another integer. If the sum of the integers is $$57$$, find the integers.
6. A larger integer is $$5$$ more than twice another integer. If the sum of the integers is $$83$$, find the integers.
7. One integer is $$3$$ less than twice another integer. Find the integers if their sum is $$135$$.
8. One integer is $$10$$ less than $$4$$ times another integer. Find the integers if their sum is $$100$$.
9. The sum of three consecutive integers is $$339$$. Find the integers.
10. The sum of four consecutive integers is $$130$$. Find the integers.
11. The sum of three consecutive even integers is $$174$$. Find the integers.
12. The sum of four consecutive even integers is $$116$$. Find the integers.
13. The sum of three consecutive odd integers is $$81$$. Find the integers.
14. The sum of four consecutive odd integers is $$176$$. Find the integers.

1. $$−5$$

3. $$9$$

5. $$14, 43$$

7. $$46, 89$$

9. $$112, 113, 114$$

11. $$56, 58, 60$$

13. $$25, 27, 29$$

Exercise $$\PageIndex{7}$$

Geometry Problems

1. The length of a rectangle is $$5$$ centimeters less than twice its width. If the perimeter is $$134$$ centimeters, find the length and width.
2. The length of a rectangle is $$4$$ centimeters more than $$3$$ times its width. If the perimeter is $$64$$ centimeters, find the length and width.
3. The width of a rectangle is one-half that of its length. If the perimeter measures $$36$$ inches, find the dimensions of the rectangle.
4. The width of a rectangle is $$4$$ inches less than its length. If the perimeter measures $$72$$ inches, find the dimensions of the rectangle.
5. The perimeter of a square is $$48$$ inches. Find the length of each side.
6. The perimeter of an equilateral triangle is $$96$$ inches. Find the length of each side.
7. The circumference of a circle measures $$80π$$ units. Find the radius.
8. The circumference of a circle measures $$25$$ centimeters. Find the radius rounded off to the nearest hundredth.

1. Width: $$24$$ centimeters; length: $$43$$ centimeters

3. Width: $$6$$ inches; length: $$12$$ inches

5. $$12$$ inches

7. $$40$$ units

Exercise $$\PageIndex{8}$$

Simple Interest Problems

1. For how many years must $$1,000$$ be invested at $$5\frac{1}{2}$$ % to earn $$165$$ in simple interest?
2. For how many years must $$20,000$$ be invested at $$6\frac{1}{4}$$ % to earn $$3,125$$ in simple interest?
3. At what annual interest rate must $$6500$$ be invested for $$2$$ years to yield $$1,040$$ in simple interest?
4. At what annual interest rate must $$5,750$$ be invested for $$1$$ year to yield $$333.50$$ in simple interest?
5. If the simple interest earned for $$5$$ years was $$1,860$$ and the annual interest rate was $$6$$%, what was the principal?
6. If the simple interest earned for $$2$$ years was $$543.75$$ and the annual interest rate was $$3\frac{3}{4}$$ %, what was the principal?
7. How many years will it take $$600$$ to double earning simple interest at a $$5$$% annual rate? (Hint: To double, the investment must earn $$600$$ in simple interest.)
8. How many years will it take $$10,000$$ to double earning simple interest at a $$5$$% annual rate? (Hint: To double, the investment must earn $$10,000$$ in simple interest.)
9. Jim invested $$4,200$$ in two accounts. One account earns $$3$$% simple interest and the other earns $$6$$%. If the interest after $$1$$ year was $$159$$, how much did he invest in each account?
10. Jane has her $$6,500$$ savings invested in two accounts. She has part of it in a CD at $$5$$% annual interest and the rest in a savings account that earns $$4$$% annual interest. If the simple interest earned from both accounts is $$303$$ for the year, then how much does she have in each account?
11. Jose put last year’s bonus of $$8,400$$ into two accounts. He invested part in a CD with $$2.5$$% annual interest and the rest in a money market fund with $$1.5$$% annual interest. His total interest for the year was $$198$$. How much did he invest in each account?
12. Mary invested her total savings of $$3,300$$ in two accounts. Her mutual fund account earned $$6.2$$% last year and her CD earned $$2.4$$%. If her total interest for the year was $$124.80$$, how much was in each account?
13. Alice invests money into two accounts, one with $$3$$% annual interest and another with $$5$$% annual interest. She invests $$3$$ times as much in the higher yielding account as she does in the lower yielding account. If her total interest for the year is $$126$$, how much did she invest in each account?
14. James invested an inheritance in two separate banks. One bank offered $$5\frac{1}{2}$$ % annual interest rate and the other $$6\frac{1}{4}$$%. He invested twice as much in the higher yielding bank account than he did in the other. If his total simple interest for $$1$$ year was $$5,760$$, then what was the amount of his inheritance?

1. $$3$$ years

3. $$8$$%

5. $$6,200$$

7. $$20$$ years

9. He invested $$3,100$$ at $$3$$% and $$1,100$$ at $$6$$%.

11. Jose invested $$7,200$$ in the CD and $$1,200$$ in the money market fund.

13. Alice invested $$700$$ at $$3$$% and $$2,100$$ at $$5$$%.

Exercise $$\PageIndex{9}$$

Uniform Motion Problems

1. If it takes Jim $$1 \frac{1}{4}$$ hours to drive the $$40$$ miles to work, then what is Jim’s average speed?
2. It took Jill $$3 \frac{1}{2}$$ hours to drive the $$189$$ miles home from college. What was her average speed?
3. At what speed should Jim drive if he wishes to travel $$176$$ miles in $$2 \frac{3}{4}$$ hours?
4. James and Martin were able to drive the $$1,140$$ miles from Los Angeles to Seattle. If the total trip took $$19$$ hours, then what was their average speed?

1. $$32$$ miles per hour

3. $$64$$ miles per hour

Exercise $$\PageIndex{10}$$

1. What is regarded as the main business of algebra? Explain.
2. What is the origin of the word algebra?
3. Create an identity or contradiction of your own and share it on the discussion board. Provide a solution and explain how you found it.
4. Post something you found particularly useful or interesting in this section. Explain why.
5. Conduct a web search for “solving linear equations.” Share a link to website or video tutorial that you think is helpful.

Footnotes

129Statement indicating that two algebraic expressions are equal.

130An equation that can be written in the standard form $$ax + b = 0$$, where $$a$$ and $$b$$ are real numbers and $$a ≠ 0$$.

131Any value that can replace the variable in an equation to produce a true statement.

132Equations with the same solution set.

133Properties that allow us to obtain equivalent equations by adding, subtracting, multiplying, and dividing both sides of an equation by nonzero real numbers.

134Allows you to solve for the variable on either side of the equal sign, because $$x = 5$$ is equivalent to $$5 = x$$.

135Equations that are true for particular values.

136An equation that is true for all possible values.

137An equation that is never true and has no solution.