Solve 2(x-3)+xleft(x-13right)<left(x-2right)^2-7x

$$2(x-3)+x \left( x-13 \right) < { \left(x-2 \right) }^{ 2 } -7x$$

Solve for x

$$x\in \mathrm{R}$$

Solution Steps

Use the distributive property to multiply 2 by x-3.

$$2x-6+x\left(x-13\right)<\left(x-2\right)^{2}-7x$$

Use the distributive property to multiply x by x-13.

$$2x-6+x^{2}-13x<\left(x-2\right)^{2}-7x$$

Combine 2x and -13x to get -11x.

$$-11x-6+x^{2}<\left(x-2\right)^{2}-7x$$

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

$$-11x-6+x^{2}<x^{2}-4x+4-7x$$

Combine -4x and -7x to get -11x.

$$-11x-6+x^{2}<x^{2}-11x+4$$

Subtract x^{2} from both sides.

$$-11x-6+x^{2}-x^{2}<-11x+4$$

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

$$-11x-6<-11x+4$$

Add 11x to both sides.

$$-11x-6+11x<4$$

Combine -11x and 11x to get 0.

$$-6<4$$

This is true for any x.

$$x\in \mathrm{R}$$