Solve left(x-3right)left(x+2right)-left(x-3right)^2geq15x-10

$$\left( x-3 \right) \left( x+2 \right) - { \left(x-3 \right) }^{ 2 } \geq 15x-10$$

Solve for x

$$x\leq -\frac{1}{2}$$

Solution Steps

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

$$x^{2}-x-6-\left(x-3\right)^{2}\geq 15x-10$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-3\right)^{2}.

$$x^{2}-x-6-\left(x^{2}-6x+9\right)\geq 15x-10$$

To find the opposite of x^{2}-6x+9, find the opposite of each term.

$$x^{2}-x-6-x^{2}+6x-9\geq 15x-10$$

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

$$-x-6+6x-9\geq 15x-10$$

Combine -x and 6x to get 5x.

$$5x-6-9\geq 15x-10$$

Subtract 9 from -6 to get -15.

$$5x-15\geq 15x-10$$

Subtract 15x from both sides.

$$5x-15-15x\geq -10$$

Combine 5x and -15x to get -10x.

$$-10x-15\geq -10$$

Add 15 to both sides.

$$-10x\geq -10+15$$

Add -10 and 15 to get 5.

$$-10x\geq 5$$

Divide both sides by -10. Since -10 is negative, the inequality direction is changed.

$$x\leq \frac{5}{-10}$$

Reduce the fraction \frac{5}{-10} to lowest terms by extracting and canceling out 5.

$$x\leq -\frac{1}{2}$$