Algebra Examples

$- 4 x + 5 < - x - 7 < - 4 x + 8$

Step 1

Move all terms not containing $x$ from the center section of the inequality.

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Step 1.1

Add $7$ to each section of the inequality because it does not contain the variable we are trying to solve for.

$- 4 x + 5 + 7 < - x < - 4 x + 8 + 7$

Step 1.2

Add $5$ and $7$ .

$- 4 x + 12 < - x < - 4 x + 8 + 7$

Step 1.3

Add $8$ and $7$ .

$- 4 x + 12 < - x < - 4 x + 15$

Step 2

Multiply each term in the inequality by $- 1$ .

$\frac{- 4 x}{- 1} + \frac{12}{- 1} > \frac{- x}{- 1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 3

Simplify each term.

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Step 3.1

Move the negative one from the denominator of $\frac{- 4 x}{- 1}$ .

$- 1 \cdot (- 4 x) + \frac{12}{- 1} > \frac{- x}{- 1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 3.2

Rewrite $- 1 \cdot (- 4 x)$ as $- (- 4 x)$ .

$- (- 4 x) + \frac{12}{- 1} > \frac{- x}{- 1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 3.3

Multiply $- 4$ by $- 1$ .

$4 x + \frac{12}{- 1} > \frac{- x}{- 1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 3.4

Divide $12$ by $- 1$ .

$4 x - 12 > \frac{- x}{- 1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 4

Dividing two negative values results in a positive value.

$4 x - 12 > \frac{x}{1} > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 5

Divide $x$ by $1$ .

$4 x - 12 > x > \frac{- 4 x}{- 1} + \frac{15}{- 1}$

Step 6

Simplify each term.

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Step 6.1

Move the negative one from the denominator of $\frac{- 4 x}{- 1}$ .

$4 x - 12 > x > - 1 \cdot (- 4 x) + \frac{15}{- 1}$

Step 6.2

Rewrite $- 1 \cdot (- 4 x)$ as $- (- 4 x)$ .

$4 x - 12 > x > - (- 4 x) + \frac{15}{- 1}$

Step 6.3

Multiply $- 4$ by $- 1$ .

$4 x - 12 > x > 4 x + \frac{15}{- 1}$

Step 6.4

Divide $15$ by $- 1$ .

$4 x - 12 > x > 4 x - 15$

Step 7

Convert the inequality to interval notation.

$(4 x - 12, 4 x - 15)$

Step 8