Solve Linear Inequalities

Graph Inequalities on the Number Line

Do you remember what it means for a number to be a solution to an equation? A solution of an equation is a value of a variable that makes a true statement when substituted into the equation.

What about the solution of an inequality? What number would make the inequality $x > 3$ true? Are you thinking, ‘x could be 4’? That’s correct, but x could be 5 too, or 20, or even 3.001. Any number greater than 3 is a solution to the inequality $x > 3$ .

We show the solutions to the inequality $x > 3$ on the number line by shading in all the numbers to the right of 3, to show that all numbers greater than 3 are solutions. Because the number 3 itself is not a solution, we put an open parenthesis at 3. The graph of $x > 3$ is shown in Figure 1. Please note that the following convention is used: thick arrows point in the positive direction and thin arrows point in the negative direction.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than 3 is graphed on the number line, with an open parenthesis at x equals 3, and a red line extending to the right of the parenthesis. — The inequality $x > 3$ is graphed on this number line.

The graph of the inequality $x \geq 3$ is very much like the graph of $x > 3$ , but now we need to show that 3 is a solution, too. We do that by putting a bracket at $x = 3$ , as shown in Figure 2.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is greater than or equal to 3 is graphed on the number line, with an open bracket at x equals 3, and a red line extending to the right of the bracket. — The inequality $x \geq 3$ is graphed on this number line.

Notice that the open parentheses symbol, (, shows that the endpoint of the inequality is not included. The open bracket symbol, [, shows that the endpoint is included.

Graph on the number line:

ⓐ $x \leq 1$ ⓑ $x < 5$ ⓒ $x > - 1$

Solution

ⓐ $x \leq 1$
This means all numbers less than or equal to 1. We shade in all the numbers on the number line to the left of 1 and put a bracket at $x = 1$ to show that it is included.
ⓑ $x < 5$
This means all numbers less than 5, but not including 5. We shade in all the numbers on the number line to the left of 5 and put a parenthesis at $x = 5$ to show it is not included.
ⓒ $x > - 1$
This means all numbers greater than $−1$ , but not including $−1$ . We shade in all the numbers on the number line to the right of $−1$ , then put a parenthesis at $x = −1$ to show it is not included.

We can also represent inequalities using interval notation. As we saw above, the inequality $x > 3$ means all numbers greater than 3. There is no upper end to the solution to this inequality. In interval notation, we express $x > 3$ as $(3, \infty) .$ The symbol $\infty$ is read as ‘infinity’. It is not an actual number. Figure 3 shows both the number line and the interval notation.

The inequality $x \leq 1$ means all numbers less than or equal to 1. There is no lower end to those numbers. We write $x \leq 1$ in interval notation as $(− \infty, 1]$ . The symbol $− \infty$ is read as ‘negative infinity’. Figure 4 shows both the number line and interval notation.

This figure is a number line ranging from negative 5 to 5 with tick marks for each integer. The inequality x is less than or equal to 1 is graphed on the number line, with an open bracket at x equals 1, and a red line extending to the left of the bracket. The inequality is also written in interval notation as parenthesis, negative infinity comma 1, bracket. — The inequality $x \leq 1$ is graphed on this number line and written in interval notation.

Did you notice how the parenthesis or bracket in the interval notation matches the symbol at the endpoint of the arrow? These relationships are shown in Figure 5.

This figure shows the same four number lines as above, with the same interval notation labels. Below the interval notation for each number line, there is text indicating how the notation on the number lines is similar to the interval notation. The first number line is a graph of x is greater than a, and the interval notation is parenthesis, a comma infinity, parenthesis. The text below reads: “Both have a left parenthesis.” The second number line is a graph of x is greater than or equal to a, and the interval notation is bracket, a comma infinity, parenthesis. The text below reads: “Both have a left bracket.” The third number line is a graph of x is less than a, and the interval notation is parenthesis, negative infinity comma a, parenthesis. The text below reads: “Both have a right parenthesis.” The last number line is a graph of x is less than or equal to a, and the interval notation is parenthesis, negative infinity comma a, bracket. The text below reads: “Both have a right bracket.” — The notation for inequalities on a number line and in interval notation use similar symbols to express the endpoints of intervals.

Graph on the number line and write in interval notation.

ⓐ $x \geq −3$ ⓑ $x < 2.5$ ⓒ $x \leq - \frac{3}{5}$

Solution

ⓐ

Shade to the right of $−3$ , and put a bracket at $−3$ .

Write in interval notation.
ⓑ

Shade to the left of $2.5$ , and put a parenthesis at $2.5$ .

Write in interval notation.
ⓒ

Shade to the left of $- \frac{3}{5}$ , and put a bracket at $- \frac{3}{5}$ .

Write in interval notation.

Solve Inequalities using the Subtraction and Addition Properties of Inequality

The Subtraction and Addition Properties of Equality state that if two quantities are equal, when we add or subtract the same amount from both quantities, the results will be equal.

Similar properties hold true for inequalities.

For example, we know that −4 is less than 2.
If we subtract 5 from both quantities, is the left side still less than the right side?
We get −9 on the left and −3 on the right.
And we know −9 is less than −3.
	The inequality sign stayed the same.

Similarly we could show that the inequality also stays the same for addition.

This leads us to the Subtraction and Addition Properties of Inequality.

We use these properties to solve inequalities, taking the same steps we used to solve equations. Solving the inequality $x + 5 > 9$ , the steps would look like this:

Steps to solve the inequality x + 5 > 9, demonstrating the process of isolating x.
	$x + 5 > 9$
Subtract 5 from both sides to isolate $x$ .	$x + 5 - 5 > 9 - 5$
Simplify.	$x > 4$

Any number greater than 4 is a solution to this inequality.

Solve the inequality $n - \frac{1}{2} \leq \frac{5}{8}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Add $\frac{1}{2}$ to both sides of the inequality.
Simplify.	$A mathematical inequality shows 'n' is less than or equal to the fraction 9/8, presented in black font on a white background.$
Graph the solution on the number line.
Write the solution in interval notation.	$(- \infty, \frac{9}{8}]$

Solve Inequalities using the Division and Multiplication Properties of Inequality

The Division and Multiplication Properties of Equality state that if two quantities are equal, when we divide or multiply both quantities by the same amount, the results will also be equal (provided we don’t divide by 0).

Are there similar properties for inequalities? What happens to an inequality when we divide or multiply both sides by a constant?

Consider some numerical examples.


Divide both sides by 5.	$A math problem showing two fractions, 10/5 and 15/5, with a red question mark between them, implying a comparison or operation needs to be determined.$	Multiply both sides by 5.
Simplify.
Fill in the inequality signs.

The inequality signs stayed the same.

Does the inequality stay the same when we divide or multiply by a negative number?


Divide both sides by −5.	$A mathematical problem comparing two fractions, 10/-5 and 15/-5, with a red question mark positioned between them, indicating the need for a relational operator.$	Multiply both sides by −5.
Simplify.
Fill in the inequality signs.

The inequality signs reversed their direction.

When we divide or multiply an inequality by a positive number, the inequality sign stays the same. When we divide or multiply an inequality by a negative number, the inequality sign reverses.

Here are the Division and Multiplication Properties of Inequality for easy reference.

When we divide or multiply an inequality by a:

positive number, the inequality stays the same.
negative number, the inequality reverses.

Solve the inequality $7 y < 42$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by 7. Since $7 > 0$ , the inequality stays the same.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $−10 a \geq 50$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Divide both sides of the inequality by −10. Since $−10 < 0$ , the inequality reverses.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $−20 < \frac{4}{5} u$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $\frac{5}{4}$ . Since $\frac{5}{4} > 0$ , the inequality stays the same.
Simplify.
Rewrite the variable on the left.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $\frac{t}{−2} \geq 8$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides of the inequality by $−2$ . Since $−2 < 0$ , the inequality reverses.	$An algebraic inequality is displayed, showing -2 multiplied by the fraction t/-2, which is less than or equal to -2 multiplied by 8.$
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve Inequalities That Require Simplification

Most inequalities will take more than one step to solve. We follow the same steps we used in the general strategy for solving linear equations, but be sure to pay close attention during multiplication or division.

Solve the inequality $4 m \leq 9 m + 17$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Subtract $9 m$ from both sides to collect the variables on the left.
Simplify.
Divide both sides of the inequality by −5, and reverse the inequality.
Simplify.
Graph the solution on the number line.
Write the solution in interval notation.

Solve the inequality $8 p + 3 (p - 12) > 7 p - 28$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8 p + 3 (p - 12) > 7 p - 28$
Distribute.	$8 p + 3 p - 36 > 7 p - 28$
Combine like terms.	$11 p - 36 > 7 p - 28$
Subtract $7 p$ from both sides to collect the variables on the left.	$11 p - 36 - 7 p > 7 p - 28 - 7 p$
Simplify.	$4 p - 36 > −28$
Add 36 to both sides to collect the constants on the right.	$4 p - 36 + 36 > −28 + 36$
Simplify.	$4 p > 8$
Divide both sides of the inequality by 4; the inequality stays the same.	$\frac{4 p}{4} > \frac{8}{4}$
Simplify.	$p > 2$
Graph the solution on the number line.
Write the solution in interal notation.	$(2, ∞)$

Just like some equations are identities and some are contradictions, inequalities may be identities or contradictions, too. We recognize these forms when we are left with only constants as we solve the inequality. If the result is a true statement, we have an identity. If the result is a false statement, we have a contradiction.

Solve the inequality $8 x - 2 (5 - x) < 4 (x + 9) + 6 x$ , graph the solution on the number line, and write the solution in interval notation.

Solution

Simplify each side as much as possible.	$8 x - 2 (5 - x) < 4 (x + 9) + 6 x$
Distribute.	$8 x - 10 + 2 x < 4 x + 36 + 6 x$
Combine like terms.	$10 x - 10 < 10 x + 36$
Subtract $10 x$ from both sides to collect the variables on the left.	$10 x - 10 - 10 x < 10 x + 36 - 10 x$
Simplify.	$−10 < 36$
The $x$ ’s are gone, and we have a true statement.	The inequality is an identity. The solution is all real numbers.
Graph the solution on the number line.
Write the solution in interval notation.	$(−∞, ∞)$

Solve the inequality $\frac{1}{3} a - \frac{1}{8} a > \frac{5}{24} a + \frac{3}{4}$ , graph the solution on the number line, and write the solution in interval notation.

Solution


Multiply both sides by the LCD, 24, to clear the fractions.
Simplify.
Combine like terms.
Subtract $5 a$ from both sides to collect the variables on the left.
Simplify.
The statement is false!	The inequality is a contradiction.
	There is no solution.
Graph the solution on the number line.
Write the solution in interval notation.	There is no solution.

Translate to an Inequality and Solve

To translate English sentences into inequalities, we need to recognize the phrases that indicate the inequality. Some words are easy, like ‘more than’ and ‘less than’. But others are not as obvious.

Think about the phrase ‘at least’ – what does it mean to be ‘at least 21 years old’? It means 21 or more. The phrase ‘at least’ is the same as ‘greater than or equal to’.

Table 17 shows some common phrases that indicate inequalities.

$>$	$\geq$	$<$	$\leq$
is greater than	is greater than or equal to	is less than	is less than or equal to
is more than	is at least	is smaller than	is at most
is larger than	is no less than	has fewer than	is no more than
exceeds	is the minimum	is lower than	is the maximum

Translate and solve. Then write the solution in interval notation and graph on the number line.

Twelve times c is no more than 96.

Solution

Translate.
Solve—divide both sides by 12.
Simplify.
Write in interval notation.
Graph on the number line.

Translate and solve. Then write the solution in interval notation and graph on the number line.

Thirty less than x is at least 45.

Solution

Translate.
Solve—add 30 to both sides.
Simplify.
Write in interval notation.
Graph on the number line.

Key Concepts

Subtraction Property of Inequality
For any numbers a, b, and c,
if $a < b$ then $a - c < b - c$ and
if $a > b$ then $a - c > b - c .$
Addition Property of Inequality
For any numbers a, b, and c,
if $a < b$ then $a + c < b + c$ and
if $a > b$ then $a + c > b + c .$
Division and Multiplication Properties of Inequality
For any numbers a, b, and c,
if $a < b$ and $c > 0$ , then $\frac{a}{c} < \frac{b}{c}$ and $a c > b c$ .
if $a > b$ and $c > 0$ , then $\frac{a}{c} > \frac{b}{c}$ and $a c > b c$ .
if $a < b$ and $c < 0$ , then $\frac{a}{c} > \frac{b}{c}$ and $a c > b c$ .
if $a > b$ and $c < 0$ , then $\frac{a}{c} < \frac{b}{c}$ and $a c < b c$ .
When we divide or multiply an inequality by a:
- positive number, the inequality stays the same.
- negative number, the inequality reverses.

Section Exercises

Practice Makes Perfect

Graph Inequalities on the Number Line

In the following exercises, graph each inequality on the number line.

ⓐ $x \leq - 2$
ⓑ $x > - 1$ ⓒ $x < 0$

ⓐ $x > 1$ ⓑ $x < - 2$
ⓒ $x \geq −3$

Solution

ⓐ $x \geq −3$ ⓑ $x < 4$ ⓒ $x \leq - 2$

ⓐ $x \leq 0$ ⓑ $x > - 4$ ⓒ $x \geq −1$

Solution

In the following exercises, graph each inequality on the number line and write in interval notation.

ⓐ $x < - 2$ ⓑ $x \geq −3.5$ ⓒ $x \leq \frac{2}{3}$

ⓐ $x > 3$ ⓑ $x \leq - 0.5$ ⓒ $x \geq \frac{1}{3}$

Solution

ⓐ $x \geq −4$ ⓑ $x < 2.5$ ⓒ $x > - \frac{3}{2}$

ⓐ $x \leq 5$ ⓑ $x \geq −1.5$ ⓒ $x < - \frac{7}{3}$

Solution

Solve Inequalities using the Subtraction and Addition Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$n - 11 < 33$

$m - 45 \leq 62$

Solution

$u + 25 > 21$

$v + 12 > 3$

Solution

$a + \frac{3}{4} \geq \frac{7}{10}$

$b + \frac{7}{8} \geq \frac{1}{6}$

Solution

$f - \frac{13}{20} < - \frac{5}{12}$

$g - \frac{11}{12} < - \frac{5}{18}$

Solution

Solve Inequalities using the Division and Multiplication Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$8 x > 72$

$6 y < 48$

Solution

$7 r \leq 56$

$9 s \geq 81$

Solution

$−5 u \geq 65$

$−8 v \leq 96$

Solution

$−9 c < 126$

$−7 d > 105$

Solution

$20 > \frac{2}{5} h$

$40 < \frac{5}{8} k$

Solution

$\frac{7}{6} j \geq 42$

$\frac{9}{4} g \leq 36$

Solution

$\frac{a}{−3} \leq 9$

$\frac{b}{−10} \geq 30$

Solution

$−25 < \frac{p}{−5}$

$−18 > \frac{q}{−6}$

Solution

$9 t \geq −27$

$7 s < - 28$

Solution

$\frac{2}{3} y > - 36$

$\frac{3}{5} x \leq - 45$

Solution

Solve Inequalities That Require Simplification

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$4 v \geq 9 v - 40$

$5 u \leq 8 u - 21$

Solution

$13 q < 7 q - 29$

$9 p > 14 p - 18$

Solution

$12 x + 3 (x + 7) > 10 x - 24$

$9 y + 5 (y + 3) < 4 y - 35$

Solution

$6 h - 4 (h - 1) \leq 7 h - 11$

$4 k - (k - 2) \geq 7 k - 26$

Solution

$8 m - 2 (14 - m) \geq 7 (m - 4) + 3 m$

$6 n - 12 (3 - n) \leq 9 (n - 4) + 9 n$

Solution

$\frac{3}{4} b - \frac{1}{3} b < \frac{5}{12} b - \frac{1}{2}$

$9 u + 5 (2 u - 5) \geq 12 (u - 1) + 7 u$

Solution

At the top of this figure is the result of the inequality: the inequality is a contradiction. Below this is a number line ranging from negative 2 to 2 with tick marks for each integer. Because this is a contradiction, no inequality is graphed on the number line. Below the number line is the statement: “No solution”.

$\frac{2}{3} g - \frac{1}{2} (g - 14) \leq \frac{1}{6} (g + 42)$

$\frac{5}{6} a - \frac{1}{4} a > \frac{7}{12} a + \frac{2}{3}$

Solution

$\frac{4}{5} h - \frac{2}{3} (h - 9) \geq \frac{1}{15} (2 h + 90)$

$12 v + 3 (4 v - 1) \leq 19 (v - 2) + 5 v$

Solution

Mixed practice

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$15 k \leq - 40$

$35 k \geq −77$

Solution

$23 p - 2 (6 - 5 p) > 3 (11 p - 4)$

$18 q - 4 (10 - 3 q) < 5 (6 q - 8)$

Solution

$- \frac{9}{4} x \geq - \frac{5}{12}$

$- \frac{21}{8} y \leq - \frac{15}{28}$

Solution

$c + 34 < - 99$

$d + 29 > - 61$

Solution

$\frac{m}{18} \geq −4$

$\frac{n}{13} \leq - 6$

Solution

Translate to an Inequality and Solve

In the following exercises, translate and solve .Then write the solution in interval notation and graph on the number line.

Fourteen times d is greater than 56.

Ninety times c is less than 450.

Solution

Eight times z is smaller than $−40$ .

Ten times y is at most $−110$ .

Solution

Three more than h is no less than 25.

Six more than k exceeds 25.

Solution

Ten less than w is at least 39.

Twelve less than x is no less than 21.

Solution

Negative five times r is no more than 95.

Negative two times s is lower than 56.

Solution

Nineteen less than b is at most $−22$ .

Fifteen less than a is at least $−7$ .

Solution

Everyday Math

Safety A child’s height, h, must be at least 57 inches for the child to safely ride in the front seat of a car. Write this as an inequality.

Fighter pilots The maximum height, h, of a fighter pilot is 77 inches. Write this as an inequality.

Solution

$h \leq 77$

Elevators The total weight, w, of an elevator’s passengers can be no more than 1,200 pounds. Write this as an inequality.

Shopping The number of items, n, a shopper can have in the express check-out lane is at most 8. Write this as an inequality.

Solution

$n \leq 8$

Writing Exercises

Give an example from your life using the phrase ‘at least’.

Give an example from your life using the phrase ‘at most’.

Solution

Answers will vary.

Explain why it is necessary to reverse the inequality when solving $−5 x > 10$ .

Explain why it is necessary to reverse the inequality when solving $\frac{n}{−3} < 12$ .

Solution

Answers will vary.

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This is a table that has six rows and four columns. In the first row, which is a header row, the cells read from left to right: “I can…,” “confidently,” “with some help,” and “no-I don’t get it!” The first column below “I can…” reads “graph inequalities on the number line,” “solve inequalitites using the Subtraction and Addition Properties of Inequality,” “solve inequalitites using the Division and Multiplication Properties of Inequality,” “solve inequalities that require simplification,” and “translate to an inequality and solve.” The rest of the cells are blank.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

Chapter 2 Review Exercises

Solve Equations using the Subtraction and Addition Properties of Equality

Verify a Solution of an Equation

In the following exercises, determine whether each number is a solution to the equation.

$10 x - 1 = 5 x; x = \frac{1}{5}$

$w + 2 = \frac{5}{8}; w = \frac{3}{8}$

Solution

$−12 n + 5 = 8 n; n = - \frac{5}{4}$

$6 a - 3 = −7 a, a = \frac{3}{13}$

Solution

yes

Solve Equations using the Subtraction and Addition Properties of Equality

In the following exercises, solve each equation using the Subtraction Property of Equality.

$x + 7 = 19$

$y + 2 = −6$

Solution

$y = −8$

$a + \frac{1}{3} = \frac{5}{3}$

$n + 3.6 = 5.1$

Solution

$n = 1.5$

In the following exercises, solve each equation using the Addition Property of Equality.

$u - 7 = 10$

$x - 9 = −4$

Solution

$x = 5$

$c - \frac{3}{11} = \frac{9}{11}$

$p - 4.8 = 14$

Solution

$p = 18.8$

In the following exercises, solve each equation.

$n - 12 = 32$

$y + 16 = −9$

Solution

$y = −25$

$f + \frac{2}{3} = 4$

$d - 3.9 = 8.2$

Solution

$d = 12.1$

Solve Equations That Require Simplification

In the following exercises, solve each equation.

$y + 8 - 15 = −3$

$7 x + 10 - 6 x + 3 = 5$

Solution

$x = −8$

$6 (n - 1) - 5 n = −14$

$8 (3 p + 5) - 23 (p - 1) = 35$

Solution

$p = −28$

Translate to an Equation and Solve

In the following exercises, translate each English sentence into an algebraic equation and then solve it.

The sum of $−6$ and $m$ is 25.

Four less than $n$ is 13.

Solution

$n - 4 = 13; n = 17$

Translate and Solve Applications

In the following exercises, translate into an algebraic equation and solve.

Rochelle’s daughter is 11 years old. Her son is 3 years younger. How old is her son?

Tan weighs 146 pounds. Minh weighs 15 pounds more than Tan. How much does Minh weigh?

Solution

161 pounds

Peter paid $9.75 to go to the movies, which was $46.25 less than he paid to go to a concert. How much did he pay for the concert?

Elissa earned $152.84 this week, which was $21.65 more than she earned last week. How much did she earn last week?

Solution

$131.19

Solve Equations using the Division and Multiplication Properties of Equality

Solve Equations Using the Division and Multiplication Properties of Equality

In the following exercises, solve each equation using the division and multiplication properties of equality and check the solution.

$8 x = 72$

$13 a = −65$

Solution

$a = −5$

$0.25 p = 5.25$

$− y = 4$

Solution

$y = −4$

$\frac{n}{6} = 18$

$\frac{y}{−10} = 30$

Solution

$y = −300$

$36 = \frac{3}{4} x$

$\frac{5}{8} u = \frac{15}{16}$

Solution

$u = \frac{3}{2}$

$−18 m = −72$

$\frac{c}{9} = 36$

Solution

$c = 324$

$0.45 x = 6.75$

$\frac{11}{12} = \frac{2}{3} y$

Solution

$y = \frac{11}{8}$

Solve Equations That Require Simplification

In the following exercises, solve each equation requiring simplification.

$5 r - 3 r + 9 r = 35 - 2$

$24 x + 8 x - 11 x = −7 - 14$

Solution

$x = −1$

$\frac{11}{12} n - \frac{5}{6} n = 9 - 5$

$−9 (d - 2) - 15 = −24$

Solution

$d = 3$

Translate to an Equation and Solve

In the following exercises, translate to an equation and then solve.

143 is the product of $−11$ and y.

The quotient of b and 9 is $−27$ .

Solution

$\frac{b}{9} = −27; b = −243$

The sum of q and one-fourth is one.

The difference of s and one-twelfth is one fourth.

Solution

$s - \frac{1}{12} = \frac{1}{4}; s = \frac{1}{3}$

Translate and Solve Applications

In the following exercises, translate into an equation and solve.

Ray paid $21 for 12 tickets at the county fair. What was the price of each ticket?

Janet gets paid $24 per hour. She heard that this is $\frac{3}{4}$ of what Adam is paid. How much is Adam paid per hour?

Solution

$32

Solve Equations with Variables and Constants on Both Sides

Solve an Equation with Constants on Both Sides

In the following exercises, solve the following equations with constants on both sides.

$8 p + 7 = 47$

$10 w - 5 = 65$

Solution

$w = 7$

$3 x + 19 = −47$

$32 = −4 - 9 n$

Solution

$n = −4$

Solve an Equation with Variables on Both Sides

In the following exercises, solve the following equations with variables on both sides.

$7 y = 6 y - 13$

$5 a + 21 = 2 a$

Solution

$a = −7$

$k = −6 k - 35$

$4 x - \frac{3}{8} = 3 x$

Solution

$x = \frac{3}{8}$

Solve an Equation with Variables and Constants on Both Sides

In the following exercises, solve the following equations with variables and constants on both sides.

$12 x - 9 = 3 x + 45$

$5 n - 20 = −7 n - 80$

Solution

$n = −5$

$4 u + 16 = −19 - u$

$\frac{5}{8} c - 4 = \frac{3}{8} c + 4$

Solution

$c = 32$

Use a General Strategy for Solving Linear Equations

Solve Equations Using the General Strategy for Solving Linear Equations

In the following exercises, solve each linear equation.

$6 (x + 6) = 24$

$9 (2 p - 5) = 72$

Solution

$p = \frac{13}{2}$

$− (s + 4) = 18$

$8 + 3 (n - 9) = 17$

Solution

$n = 12$

$23 - 3 (y - 7) = 8$

$\frac{1}{3} (6 m + 21) = m - 7$

Solution

$m = −14$

$4 (3.5 y + 0.25) = 365$

$0.25 (q - 8) = 0.1 (q + 7)$

Solution

$q = 18$

$8 (r - 2) = 6 (r + 10)$

$5 + 7 (2 - 5 x) = 2 (9 x + 1)$
$- (13 x - 57)$

Solution

$x = −1$

$(9 n + 5) - (3 n - 7)$
$= 20 - (4 n - 2)$

$2 [−16 + 5 (8 k - 6)]$
$= 8 (3 - 4 k) - 32$

Solution

$k = \frac{3}{4}$

Classify Equations

In the following exercises, classify each equation as a conditional equation, an identity, or a contradiction and then state the solution.

$17 y - 3 (4 - 2 y) = 11 (y - 1)$
$+ 12 y - 1$

$9 u + 32 = 15 (u - 4)$
$- 3 (2 u + 21)$

Solution

contradiction; no solution

$−8 (7 m + 4) = −6 (8 m + 9)$

$21 (c - 1) - 19 (c + 1)$
$= 2 (c - 20)$

Solution

identity; all real numbers

Solve Equations with Fractions and Decimals

Solve Equations with Fraction Coefficients

In the following exercises, solve each equation with fraction coefficients.

$\frac{2}{5} n - \frac{1}{10} = \frac{7}{10}$

$\frac{1}{3} x + \frac{1}{5} x = 8$

Solution

$x = 15$

$\frac{3}{4} a - \frac{1}{3} = \frac{1}{2} a - \frac{5}{6}$

$\frac{1}{2} (k - 3) = \frac{1}{3} (k + 16)$

Solution

$k = 41$

$\frac{3 x - 2}{5} = \frac{3 x + 4}{8}$

$\frac{5 y - 1}{3} + 4 = \frac{−8 y + 4}{6}$

Solution

$y = −1$

Solve Equations with Decimal Coefficients

In the following exercises, solve each equation with decimal coefficients.

$0.8 x - 0.3 = 0.7 x + 0.2$

$0.36 u + 2.55 = 0.41 u + 6.8$

Solution

$u = −85$

$0.6 p - 1.9 = 0.78 p + 1.7$

$0.7 y + 2.5 = 0.95 y - 9.25$

Solution

$y = 47$

Solve a Formula for a Specific Variable

Use the Distance, Rate, and Time Formula

In the following exercises, solve.

Natalie drove for $7 \frac{1}{2}$ hours at 60 miles per hour. How much distance did she travel?

Mallory is taking the bus from St. Louis to Chicago. The distance is 300 miles and the bus travels at a steady rate of 60 miles per hour. How long will the bus ride be?

Solution

5 hours

Aaron’s friend drove him from Buffalo to Cleveland. The distance is 187 miles and the trip took 2.75 hours. How fast was Aaron’s friend driving?

Link rode his bike at a steady rate of 15 miles per hour for $2 \frac{1}{2}$ hours. How much distance did he travel?

Solution

37.5 miles

Solve a Formula for a Specific Variable

In the following exercises, solve.

Use the formula. $d = r t$ to solve for t
ⓐ when $d = 510$ and $r = 60$
ⓑ in general

Use the formula. $d = r t$ to solve for $r$
ⓐ when when $d = 451$ and $t = 5.5$
ⓑ in general

Solution

ⓐ $r = 82 mph$ ; ⓑ $r = \frac{D}{t}$

Use the formula $A = \frac{1}{2} b h$ to solve for $b$
ⓐ when $A = 390$ and $h = 26$
ⓑ in general

Use the formula $A = \frac{1}{2} b h$ to solve for $h$
ⓐ when $A = 153$ and $b = 18$
ⓑ in general

Solution

ⓐ $h = 17$ ⓑ $h = \frac{2 A}{b}$

Use the formula $I = P r t$ to solve for the principal, P for
ⓐ $I = $ 2, 501, r = 4.1 %,$
$t = 5 years$
ⓑ in general

Solve the formula $4 x + 3 y = 6$ for y
ⓐ when $x = −2$
ⓑ in general

Solution

ⓐ $y = \frac{14}{3}$ ⓑ $y = \frac{6 - 4 x}{3}$

Solve $180 = a + b + c$ for $c$ .

Solve the formula $V = L W H$ for $H$ .

Solution

$H = \frac{V}{L W}$

Solve Linear Inequalities

Graph Inequalities on the Number Line

In the following exercises, graph each inequality on the number line.

Solution

In the following exercises, graph each inequality on the number line and write in interval notation.

Solution

Solve Inequalities using the Subtraction and Addition Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$n - 12 \leq 23$

$m + 14 \leq 56$

Solution

$a + \frac{2}{3} \geq \frac{7}{12}$

$b - \frac{7}{8} \geq - \frac{1}{2}$

Solution

Solve Inequalities using the Division and Multiplication Properties of Inequality

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$9 x > 54$

$−12 d \leq 108$

Solution

$\frac{5}{6} j < - 60$

$\frac{q}{- 2} \geq - 24$

Solution

Solve Inequalities That Require Simplification

In the following exercises, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$6 p > 15 p - 30$

$9 h - 7 (h - 1) \leq 4 h - 23$

Solution

$5 n - 15 (4 - n) < 10 (n - 6) + 10 n$

$\frac{3}{8} a - \frac{1}{12} a > \frac{5}{12} a + \frac{3}{4}$

Solution

Translate to an Inequality and Solve

In the following exercises, translate and solve. Then write the solution in interval notation and graph on the number line.

Five more than z is at most 19.

Three less than c is at least 360.

Solution

Nine times n exceeds 42.

Negative two times a is no more than 8.

Solution

Everyday Math

Describe how you have used two topics from this chapter in your life outside of your math class during the past month.

Chapter 2 Practice Test

Determine whether each number is a solution to the equation $6 x - 3 = x + 20$ .

ⓐ 5ⓑ $\frac{23}{5}$

Solution

ⓐ no ⓑ yes

In the following exercises, solve each equation.

$n - \frac{2}{3} = \frac{1}{4}$

$\frac{9}{2} c = 144$

Solution

$c = 32$

$4 y - 8 = 16$

$−8 x - 15 + 9 x - 1 = −21$

Solution

$x = −5$

$−15 a = 120$

$\frac{2}{3} x = 6$

Solution

$x = 9$

$x - 3.8 = 8.2$

$10 y = −5 y - 60$

Solution

$y = −4$

$8 n - 2 = 6 n - 12$

$9 m - 2 - 4 m - m = 42 - 8$

Solution

$m = 9$

$−5 (2 x - 1) = 45$

$− (d - 9) = 23$

Solution

$d = −14$

$\frac{1}{4} (12 m - 28) = 6 - 2 (3 m - 1)$

$2 (6 x - 5) - 8 = −22$

Solution

$x = - \frac{1}{3}$

$8 (3 a - 5) - 7 (4 a - 3) = 20 - 3 a$

$\frac{1}{4} p - \frac{1}{3} = \frac{1}{2}$

Solution

$p = \frac{10}{3}$

$0.1 d + 0.25 (d + 8) = 4.1$

$14 n - 3 (4 n + 5) = −9 + 2 (n - 8)$

Solution

contradiction; no solution

$9 (3 u - 2) - 4 [6 - 8 (u - 1)]$ $= 3 (u - 2)$

Solve the formula $x - 2 y = 5$ for y
ⓐ when $x = −3$
ⓑ in general

Solution

ⓐ $y = 4$ ⓑ $y = \frac{5 - x}{2}$

In the following exercises, graph on the number line and write in interval notation.

$x \geq −3.5$

$x < \frac{11}{4}$

Solution

This figure is a number line ranging from 1 to 5 with tick marks for each integer. The inequality x is less than 11/4 is graphed on the number line, with an open parenthesis at x equals 11/4, and a dark line extending to the left of the parenthesis. Below the number line is the solution written in interval notation: parenthesis, negative infinity comma 11/4, parenthesis.

In the following exercises,, solve each inequality, graph the solution on the number line, and write the solution in interval notation.

$8 k \geq 5 k - 120$

$3 c - 10 (c - 2) < 5 c + 16$

Solution

In the following exercises, translate to an equation or inequality and solve.

4 less than twice x is 16.

Fifteen more than n is at least 48.

Solution

$n + 15 \geq 48; n \geq 33$

Samuel paid $25.82 for gas this week, which was $3.47 less than he paid last week. How much had he paid last week?

Jenna bought a coat on sale for $120, which was $\frac{2}{3}$ of the original price. What was the original price of the coat?

Solution

$120 = \frac{2}{3} p$ ; The original price was $180.

Sean took the bus from Seattle to Boise, a distance of 506 miles. If the trip took $7 \frac{2}{3}$ hours, what was the speed of the bus?


Shade to the right of $−3$ , and put a bracket at $−3$ .
Write in interval notation.


Shade to the left of $2.5$ , and put a parenthesis at $2.5$ .
Write in interval notation.


Shade to the left of $- \frac{3}{5}$ , and put a bracket at $- \frac{3}{5}$ .
Write in interval notation.