Solve the System of Inequalities 2x+8>4 and 20>=2x+8

$2 x + 8 > 4$ and $20 \geq 2 x + 8$

Step 1

Simplify the first inequality.

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

Subtract $8$ from both sides of the inequality.

$2 x > 4 - 8$ and $20 \geq 2 x + 8$

Subtract $8$ from $4$ .

$2 x > - 4$ and $20 \geq 2 x + 8$

Divide each term in $2 x > - 4$ by $2$ and simplify.

Divide each term in $2 x > - 4$ by $2$ .

$\frac{2 x}{2} > \frac{- 4}{2}$ and $20 \geq 2 x + 8$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} > \frac{- 4}{2}$ and $20 \geq 2 x + 8$

Divide $x$ by $1$ .

$x > \frac{- 4}{2}$ and $20 \geq 2 x + 8$

Simplify the right side.

Divide $- 4$ by $2$ .

$x > - 2$ and $20 \geq 2 x + 8$

Step 2

Simplify the second inequality.

Rewrite so $x$ is on the left side of the inequality.

$x > - 2$ and $2 x + 8 \leq 20$

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

Subtract $8$ from both sides of the inequality.

$x > - 2$ and $2 x \leq 20 - 8$

Subtract $8$ from $20$ .

$x > - 2$ and $2 x \leq 12$

Divide each term in $2 x \leq 12$ by $2$ and simplify.

Divide each term in $2 x \leq 12$ by $2$ .

$x > - 2$ and $\frac{2 x}{2} \leq \frac{12}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$x > - 2$ and $\frac{2 x}{2} \leq \frac{12}{2}$

Divide $x$ by $1$ .

$x > - 2$ and $x \leq \frac{12}{2}$

Simplify the right side.

Divide $12$ by $2$ .

$x > - 2$ and $x \leq 6$

Step 3

The intersection consists of the elements that are contained in both intervals.

$- 2 < x \leq 6$

Step 4

The result can be shown in multiple forms.

Inequality Form:

$- 2 < x \leq 6$

Interval Notation:

$(- 2, 6]$

Step 5

Solve the System of Inequalities 2x+8>4 and 20>=2x+8

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