Solve for x 2|5-2x|-3<=15

$2 | 5 - 2 x | - 3 \leq 15$

Step 1

Write $2 | 5 - 2 x | - 3 \leq 15$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$5 - 2 x \geq 0$

Solve the inequality.

Subtract $5$ from both sides of the inequality.

$- 2 x \geq - 5$

Divide each term in $- 2 x \geq - 5$ by $- 2$ and simplify.

Divide each term in $- 2 x \geq - 5$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} \leq \frac{- 5}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} \leq \frac{- 5}{- 2}$

Divide $x$ by $1$ .

$x \leq \frac{- 5}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x \leq \frac{5}{2}$

In the piece where $5 - 2 x$ is non-negative, remove the absolute value.

$2 (5 - 2 x) - 3 \leq 15$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$5 - 2 x < 0$

Solve the inequality.

Subtract $5$ from both sides of the inequality.

$- 2 x < - 5$

Divide each term in $- 2 x < - 5$ by $- 2$ and simplify.

Divide each term in $- 2 x < - 5$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} > \frac{- 5}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} > \frac{- 5}{- 2}$

Divide $x$ by $1$ .

$x > \frac{- 5}{- 2}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x > \frac{5}{2}$

In the piece where $5 - 2 x$ is negative, remove the absolute value and multiply by $- 1$ .

$2 (- (5 - 2 x)) - 3 \leq 15$

Write as a piecewise.

${\begin{cases} 2 (5 - 2 x) - 3 \leq 15 & x \leq \frac{5}{2} \\ 2 (- (5 - 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Simplify $2 (5 - 2 x) - 3 \leq 15$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} 2 \cdot 5 + 2 (- 2 x) - 3 \leq 15 & x \leq \frac{5}{2} \\ 2 (- (5 - 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $2$ by $5$ .

${\begin{cases} 10 + 2 (- 2 x) - 3 \leq 15 & x \leq \frac{5}{2} \\ 2 (- (5 - 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $- 2$ by $2$ .

${\begin{cases} 10 - 4 x - 3 \leq 15 & x \leq \frac{5}{2} \\ 2 (- (5 - 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Subtract $3$ from $10$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 2 (- (5 - 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Simplify $2 (- (5 - 2 x)) - 3 \leq 15$ .

Simplify each term.

Apply the distributive property.

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 2 (- 1 \cdot 5 - (- 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $- 1$ by $5$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 2 (- 5 - (- 2 x)) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $- 2$ by $- 1$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 2 (- 5 + 2 x) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Apply the distributive property.

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 2 \cdot - 5 + 2 (2 x) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $2$ by $- 5$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ - 10 + 2 (2 x) - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Multiply $2$ by $2$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ - 10 + 4 x - 3 \leq 15 & x > \frac{5}{2} \end{cases}$

Subtract $3$ from $- 10$ .

${\begin{cases} - 4 x + 7 \leq 15 & x \leq \frac{5}{2} \\ 4 x - 13 \leq 15 & x > \frac{5}{2} \end{cases}$

Step 2

Solve $- 4 x + 7 \leq 15$ when $x \leq \frac{5}{2}$ .

Solve $- 4 x + 7 \leq 15$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Subtract $7$ from both sides of the inequality.

$- 4 x \leq 15 - 7$

Subtract $7$ from $15$ .

$- 4 x \leq 8$

Divide each term in $- 4 x \leq 8$ by $- 4$ and simplify.

Divide each term in $- 4 x \leq 8$ by $- 4$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 4 x}{- 4} \geq \frac{8}{- 4}$

Simplify the left side.

Cancel the common factor of $- 4$ .

Cancel the common factor.

$\frac{- 4 x}{- 4} \geq \frac{8}{- 4}$

Divide $x$ by $1$ .

$x \geq \frac{8}{- 4}$

Simplify the right side.

Divide $8$ by $- 4$ .

$x \geq - 2$

Find the intersection of $x \geq - 2$ and $x \leq \frac{5}{2}$ .

$- 2 \leq x \leq \frac{5}{2}$

Step 3

Solve $4 x - 13 \leq 15$ when $x > \frac{5}{2}$ .

Solve $4 x - 13 \leq 15$ for $x$ .

Move all terms not containing $x$ to the right side of the inequality.

Add $13$ to both sides of the inequality.

$4 x \leq 15 + 13$

Add $15$ and $13$ .

$4 x \leq 28$

Divide each term in $4 x \leq 28$ by $4$ and simplify.

Divide each term in $4 x \leq 28$ by $4$ .

$\frac{4 x}{4} \leq \frac{28}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} \leq \frac{28}{4}$

Divide $x$ by $1$ .

$x \leq \frac{28}{4}$

Simplify the right side.

Divide $28$ by $4$ .

$x \leq 7$

Find the intersection of $x \leq 7$ and $x > \frac{5}{2}$ .

$\frac{5}{2} < x \leq 7$

Step 4

Find the union of the solutions.

$- 2 \leq x \leq 7$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$- 2 \leq x \leq 7$

Interval Notation:

$[- 2, 7]$

Step 6

Solve for x 2|5-2x|-3<=15

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