We are asked to place each inequality in the correct box. Let's begin by naming each one of the inequalities.

A) B) C) 5 x - 6 + x \geq 2 x - 8 D) x - 8 + 4 x \leq 3 (x - 3) + 2 x 2 (3 x + 8) > 3 (2 x + 6) E) 9 x - 3 < 12 or 6 x + 2 > - 10 17 < 4 x + 5 < 21 F) 5 (x - 1) \leq 5 x - 3

Let's solve each of them separately!

Inequality A

We can solve this inequality by performing inverse operations until we isolate the variable

x .

5 x - 6 + x \geq 2 x - 8

▼

Solve for

x

AddIneq

$LHS + 6 \geq RHS + 6$

5 x + x \geq 2 x - 2

AddTerms

Add terms

6 x \geq 2 x - 2

SubIneq

$LHS - 2 x \geq RHS - 2 x$

4 x \geq - 2

DivIneq

$LHS / 4 \geq RHS / 4$

x \geq - \frac{2}{4}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

x \geq - \frac{1}{2}

All numbers greater than

- \frac{1}{2}

are solutions to the inequality. Therefore, any integer greater than this number is a solution, such as

x = 1

or

x = 2 .

Inequality B

We should start solving this inequality by distributing

2

and

3

inside the parentheses.

2 (3 x + 8) > 3 (2 x + 6)

▼

Solve for

x

Distr

Distribute $2$

6 x + 16 > 3 (2 x + 6)

Distr

Distribute $3$

6 x + 16 > 6 x + 18

SubIneq

$LHS - 6 x > RHS - 6 x$

16 ≯ 18

Since

16

is never greater than

18,

there is no solution to this inequality. This includes any possible integer solutions.

Inequality C

This compound inequality can be solved by applying inverse operations on all three sides of the inequality.

17 < 4 x + 5 < 21

▼

Solve for

x

Subtract $5$ from each expression

12 < 4 x < 16

Divide each expression by $4$

3 < x < 4

This means that

x

cannot be an integer because the solution set consists of all values between two consecutive integers.

Inequality D

Let's try to isolate

x

in this inequality.

x - 8 + 4 x \leq 3 (x - 3) + 2 x

▼

Solve for

x

Distr

Distribute $3$

x - 8 + 4 x \leq 3 x - 9 + 2 x

SimpTerms

Simplify terms

5 x - 8 \leq 5 x - 9

SubEqn

$LHS - 5 x = RHS - 5 x$

- 8 ≰ - 9

Since

8

is not greater than or equal to

9,

there are no solutions to this inequality. This includes any integer solutions.

Inequality E

Since this is an or compound inequality, we will solve each of the inequalities separately and then combine the solutions.

9 x - 3 < 12

▼

Solve for

x

AddIneq

$LHS + 3 < RHS + 3$

9 x < 15

DivIneq

$LHS / 9 < RHS / 9$

x < \frac{1 5}{9}

UseCalc

Use a calculator

x < 1.66666 \dots

RoundDec

Round to $2$ decimal place(s)

x < 1.67

So far we have that

x < 1.67 .

Since the original inequality is an or inequality we can already tell that it has at least one integer solution. For example,

x = 1

satisfies the first inequality and this is enough.

Inequality F

Let's start by distributing

5

inside the parentheses.

5 (x - 1) \leq 5 x - 3

▼

Solve for

x

Distr

Distribute $5$

5 x - 5 \leq 5 x - 3

SubIneq

$LHS - 5 x \leq RHS - 5 x$

- 5 \leq - 3

FlipIneqSign

Flip inequality and change signs

5 \geq 3

Because

5

is always greater than

3,

we have reached an identity. The solution to this inequality is all real numbers. Therefore there is at least one integer solution.

Conclusion

Finally, we can fill in the table with the information we have found.

At Least One Integer Solution	No Integer Solutions
$5 x - 6 + x \geq 2 x - 8$ $9 x - 3 < 12$ or $6 x + 2 > - 10$ $5 (x - 1) \leq 5 x - 3$	$2 (3 x + 8) > 3 (2 x + 6)$ $17 < 4 x + 5 < 21$ $x - 8 + 4 x \leq 3 (x - 3) + 2 x$

Exercise