Solve 5(4x+3)geq9(x-2)-x

$$5(4x+3) \geq 9(x-2)-x$$

Solve for x

$$x\geq -\frac{11}{4}$$

Solution Steps

Use the distributive property to multiply 5 by 4x+3.

$$20x+15\geq 9\left(x-2\right)-x$$

Use the distributive property to multiply 9 by x-2.

$$20x+15\geq 9x-18-x$$

Combine 9x and -x to get 8x.

$$20x+15\geq 8x-18$$

Subtract 8x from both sides.

$$20x+15-8x\geq -18$$

Combine 20x and -8x to get 12x.

$$12x+15\geq -18$$

Subtract 15 from both sides.

$$12x\geq -18-15$$

Subtract 15 from -18 to get -33.

$$12x\geq -33$$

Divide both sides by 12. Since 12 is positive, the inequality direction remains the same.

$$x\geq \frac{-33}{12}$$

Reduce the fraction \frac{-33}{12} to lowest terms by extracting and canceling out 3.

$$x\geq -\frac{11}{4}$$