Solve for x |x+3|>x-2

$| x + 3 | > x - 2$

Write $| x + 3 | > x - 2$ as a piecewise.

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

$x + 3 \geq 0$

Subtract $3$ from both sides of the inequality.

$x \geq - 3$

In the piece where $x + 3$ is non-negative, remove the absolute value.

$x + 3 > x - 2$

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

$x + 3 < 0$

Subtract $3$ from both sides of the inequality.

$x < - 3$

In the piece where $x + 3$ is negative, remove the absolute value and multiply by $- 1$ .

$- (x + 3) > x - 2$

Write as a piecewise.

${\begin{cases} x + 3 > x - 2 & x \geq - 3 \\ - (x + 3) > x - 2 & x < - 3 \end{cases}$

Simplify $- (x + 3) > x - 2$ .

Apply the distributive property.

${\begin{cases} x + 3 > x - 2 & x \geq - 3 \\ - x - 1 \cdot 3 > x - 2 & x < - 3 \end{cases}$

Multiply $- 1$ by $3$ .

${\begin{cases} x + 3 > x - 2 & x \geq - 3 \\ - x - 3 > x - 2 & x < - 3 \end{cases}$

Solve $x + 3 > x - 2$ when $x \geq - 3$ .

Solve $x + 3 > x - 2$ for $x$ .

Move all terms containing $x$ to the left side of the inequality.

Subtract $x$ from both sides of the inequality.

$x + 3 - x > - 2$

Combine the opposite terms in $x + 3 - x$ .

Subtract $x$ from $x$ .

$0 + 3 > - 2$

Add $0$ and $3$ .

$3 > - 2$

Since $3 > - 2$ , the inequality will always be true.

Always true

Find the intersection.

$x \geq - 3$

Solve $- x - 3 > x - 2$ when $x < - 3$ .

Solve $- x - 3 > x - 2$ for $x$ .

Move all terms containing $x$ to the left side of the inequality.

Subtract $x$ from both sides of the inequality.

$- x - 3 - x > - 2$

Subtract $x$ from $- x$ .

$- 2 x - 3 > - 2$

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

Add $3$ to both sides of the inequality.

$- 2 x > - 2 + 3$

Add $- 2$ and $3$ .

$- 2 x > 1$

Divide each term in $- 2 x > 1$ by $- 2$ and simplify.

Divide each term in $- 2 x > 1$ by $- 2$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- 2 x}{- 2} < \frac{1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 x}{- 2} < \frac{1}{- 2}$

Divide $x$ by $1$ .

$x < \frac{1}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$x < - \frac{1}{2}$

Find the intersection of $x < - \frac{1}{2}$ and $x < - 3$ .

$x < - 3$

Find the union of the solutions for any value of $x$ .

All real numbers

The result can be shown in multiple forms.

All real numbers

Interval Notation:

$(- \infty, \infty)$

Solve for x |x+3|>x-2

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