Writing and Graphing One-Variable Inequalities

a

To determine if it's a solution or not,

s

should be substituted with

3

in the inequality. Then, if the inequality holds true,

s = 3

is a solution.

s + 4 > 8

Substitute

s = 3

3 + 4 > ? 8

AddTermsAdd terms

7 ≯ 8

The statement

7 > 8

says that

7

is greater than

8 .

This is a false statement, so

s = 3

is not a solution to the inequality.

b

If

3

is a solution, the inequality will hold true when

3

is substituted for

s .

If it doesn't, it's not a solution.

3 s + 10 \leq 19

Substitute

s = 3

3 \cdot 3 + 10 \leq ? 19

MultiplyMultiply

9 + 10 \leq ? 19

AddTermsAdd terms

19 \leq 19

Since the symbol

\leq

is read as less than or equal to

19 \leq 19

is a true statement. Therefore, the answer is yes,

s = 3

is a solution.

c

We'll substitute

s

with

3

and simplify the left-hand side to determine if the inequality holds true.

- 14 \geq - 2 s - 4

Substitute

s = 3

- 14 \geq ? - 2 \cdot 3 - 4

MultiplyMultiply

- 14 \geq ? - 6 - 4

SubTermsSubtract terms

- 14 ≱ - 10

When comparing two negative numbers, the one closest to zero is the greatest. This can be visualized on a number line.

This means that $- 10$ is greater than $- 14 .$ This means that the inequality doesn't hold true for $s = 3,$ and is, thus, not a solution to the inequality.

Writing and Graphing One-Variable Inequalities 1.4 - Solution