Solve x*(x-3)+2*(3-x)leq(x+1)^2-4x

$$x \cdot ( x - 3 ) + 2 \cdot ( 3 - x ) \leq ( x + 1 ) ^ { 2 } - 4 x$$

Solve for x

$$x\geq \frac{5}{3}$$

Solution Steps

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

$$x^{2}-3x+2\left(3-x\right)\leq \left(x+1\right)^{2}-4x$$

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

$$x^{2}-3x+6-2x\leq \left(x+1\right)^{2}-4x$$

Combine -3x and -2x to get -5x.

$$x^{2}-5x+6\leq \left(x+1\right)^{2}-4x$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+1\right)^{2}.

$$x^{2}-5x+6\leq x^{2}+2x+1-4x$$

Combine 2x and -4x to get -2x.

$$x^{2}-5x+6\leq x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$x^{2}-5x+6-x^{2}\leq -2x+1$$

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

$$-5x+6\leq -2x+1$$

Add 2x to both sides.

$$-5x+6+2x\leq 1$$

Combine -5x and 2x to get -3x.

$$-3x+6\leq 1$$

Subtract 6 from both sides.

$$-3x\leq 1-6$$

Subtract 6 from 1 to get -5.

$$-3x\leq -5$$

Divide both sides by -3. Since -3 is negative, the inequality direction is changed.

$$x\geq \frac{-5}{-3}$$

Fraction \frac{-5}{-3} can be simplified to \frac{5}{3} by removing the negative sign from both the numerator and the denominator.

$$x\geq \frac{5}{3}$$