Solve (8x+1)(x-7)geq(2x-3)(4x+5)

$$(8x+1)(x-7) \geq (2x-3)(4x+5)$$

Solve for x

$$x\leq \frac{8}{53}$$

Solution Steps

Use the distributive property to multiply 8x+1 by x-7 and combine like terms.

$$8x^{2}-55x-7\geq \left(2x-3\right)\left(4x+5\right)$$

Use the distributive property to multiply 2x-3 by 4x+5 and combine like terms.

$$8x^{2}-55x-7\geq 8x^{2}-2x-15$$

Subtract 8x^{2} from both sides.

$$8x^{2}-55x-7-8x^{2}\geq -2x-15$$

Combine 8x^{2} and -8x^{2} to get 0.

$$-55x-7\geq -2x-15$$

Add 2x to both sides.

$$-55x-7+2x\geq -15$$

Combine -55x and 2x to get -53x.

$$-53x-7\geq -15$$

Add 7 to both sides.

$$-53x\geq -15+7$$

Add -15 and 7 to get -8.

$$-53x\geq -8$$

Divide both sides by -53. Since -53 is negative, the inequality direction is changed.

$$x\leq \frac{-8}{-53}$$

Fraction \frac{-8}{-53} can be simplified to \frac{8}{53} by removing the negative sign from both the numerator and the denominator.

$$x\leq \frac{8}{53}$$