Solve for x 3-(2x-5)<-4(x+2)

$3 - (2 x - 5) < - 4 (x + 2)$

Step 1

Simplify $3 - (2 x - 5)$ .

Simplify each term.

Apply the distributive property.

$3 - (2 x) - - 5 < - 4 (x + 2)$

Multiply $2$ by $- 1$ .

$3 - 2 x - - 5 < - 4 (x + 2)$

Multiply $- 1$ by $- 5$ .

$3 - 2 x + 5 < - 4 (x + 2)$

Add $3$ and $5$ .

$- 2 x + 8 < - 4 (x + 2)$

Step 2

Simplify $- 4 (x + 2)$ .

Apply the distributive property.

$- 2 x + 8 < - 4 x - 4 \cdot 2$

Multiply $- 4$ by $2$ .

$- 2 x + 8 < - 4 x - 8$

Step 3

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

Add $4 x$ to both sides of the inequality.

$- 2 x + 8 + 4 x < - 8$

Add $- 2 x$ and $4 x$ .

$2 x + 8 < - 8$

Step 4

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

Subtract $8$ from both sides of the inequality.

$2 x < - 8 - 8$

Subtract $8$ from $- 8$ .

$2 x < - 16$

Step 5

Divide each term in $2 x < - 16$ by $2$ and simplify.

Divide each term in $2 x < - 16$ by $2$ .

$\frac{2 x}{2} < \frac{- 16}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} < \frac{- 16}{2}$

Divide $x$ by $1$ .

$x < \frac{- 16}{2}$

Simplify the right side.

Divide $- 16$ by $2$ .

$x < - 8$

Step 6

The result can be shown in multiple forms.

Inequality Form:

$x < - 8$

Interval Notation:

$(- \infty, - 8)$

Solve for x 3-(2x-5)<-4(x+2)

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