Solve for x |x|+7<4

$| x | + 7 < 4$

Step 1

Write $| x | + 7 < 4$ as a piecewise.

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

$x \geq 0$

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

$x + 7 < 4$

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

$x < 0$

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

$- x + 7 < 4$

Write as a piecewise.

${\begin{cases} x + 7 < 4 & x \geq 0 \\ - x + 7 < 4 & x < 0 \end{cases}$

Step 2

Solve $x + 7 < 4$ when $x \geq 0$ .

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

Subtract $7$ from both sides of the inequality.

$x < 4 - 7$

Subtract $7$ from $4$ .

$x < - 3$

Find the intersection of $x < - 3$ and $x \geq 0$ .

No solution

Step 3

Solve $- x + 7 < 4$ when $x < 0$ .

Solve $- x + 7 < 4$ for $x$ .

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

Subtract $7$ from both sides of the inequality.

$- x < 4 - 7$

Subtract $7$ from $4$ .

$- x < - 3$

Divide each term in $- x < - 3$ by $- 1$ and simplify.

Divide each term in $- x < - 3$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

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

Simplify the left side.

Dividing two negative values results in a positive value.

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

Divide $x$ by $1$ .

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

Simplify the right side.

Divide $- 3$ by $- 1$ .

$x > 3$

Find the intersection of $x > 3$ and $x < 0$ .

No solution

Step 4

Find the union of the solutions.

No solution

Solve for x |x|+7<4

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