Solve 8(x-4)+3xleq2(x-1)+10

$$8(x-4)+3x \leq 2(x-1)+10$$

Solve for x

$$x\leq \frac{40}{9}$$

Solution Steps

Use the distributive property to multiply 8 by x-4.

$$8x-32+3x\leq 2\left(x-1\right)+10$$

Combine 8x and 3x to get 11x.

$$11x-32\leq 2\left(x-1\right)+10$$

Use the distributive property to multiply 2 by x-1.

$$11x-32\leq 2x-2+10$$

Add -2 and 10 to get 8.

$$11x-32\leq 2x+8$$

Subtract 2x from both sides.

$$11x-32-2x\leq 8$$

Combine 11x and -2x to get 9x.

$$9x-32\leq 8$$

Add 32 to both sides.

$$9x\leq 8+32$$

Add 8 and 32 to get 40.

$$9x\leq 40$$

Divide both sides by 9. Since 9 is positive, the inequality direction remains the same.

$$x\leq \frac{40}{9}$$