Solve the Inequality for a 11a-4(a+2)<=8a+10

$11 a - 4 (a + 2) \leq 8 a + 10$

Step 1

Simplify $11 a - 4 (a + 2)$ .

Simplify each term.

Apply the distributive property.

$11 a - 4 a - 4 \cdot 2 \leq 8 a + 10$

Multiply $- 4$ by $2$ .

$11 a - 4 a - 8 \leq 8 a + 10$

Subtract $4 a$ from $11 a$ .

$7 a - 8 \leq 8 a + 10$

Step 2

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

Subtract $8 a$ from both sides of the inequality.

$7 a - 8 - 8 a \leq 10$

Subtract $8 a$ from $7 a$ .

$- a - 8 \leq 10$

Step 3

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

Add $8$ to both sides of the inequality.

$- a \leq 10 + 8$

Add $10$ and $8$ .

$- a \leq 18$

Step 4

Divide each term in $- a \leq 18$ by $- 1$ and simplify.

Divide each term in $- a \leq 18$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- a}{- 1} \geq \frac{18}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{a}{1} \geq \frac{18}{- 1}$

Divide $a$ by $1$ .

$a \geq \frac{18}{- 1}$

Simplify the right side.

Divide $18$ by $- 1$ .

$a \geq - 18$

Step 5

The result can be shown in multiple forms.

Inequality Form:

$a \geq - 18$

Interval Notation:

$[- 18, \infty)$

Solve the Inequality for a 11a-4(a+2)<=8a+10

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