Solve {l}{text{5-}x+(x-3y)(x+1)+y^2+xy=(x-y)^2+x-2y+10}{5+y=0}

$$\left. \begin{array} { l } { \text { 5- } x + ( x - 3 y ) ( x + 1 ) + y ^ { 2 } + x y = ( x - y ) ^ { 2 } + x - 2 y + 10 } \\ { 5 + y = 0 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=-5$$

Solution Steps

Consider the second equation. Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$y=-5$$

Consider the first equation. Insert the known values of variables into the equation.

$$x+\left(x-3\left(-5\right)\right)\left(x+1\right)+\left(-5\right)^{2}+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Multiply -3 and -5 to get 15.

$$x+\left(x+15\right)\left(x+1\right)+\left(-5\right)^{2}+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Use the distributive property to multiply x+15 by x+1 and combine like terms.

$$x+x^{2}+16x+15+\left(-5\right)^{2}+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Combine x and 16x to get 17x.

$$17x+x^{2}+15+\left(-5\right)^{2}+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Calculate -5 to the power of 2 and get 25.

$$17x+x^{2}+15+25+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Add 15 and 25 to get 40.

$$17x+x^{2}+40+x\left(-5\right)=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Combine 17x and x\left(-5\right) to get 12x.

$$12x+x^{2}+40=\left(x-\left(-5\right)\right)^{2}+x-2\left(-5\right)+10$$

Multiply -1 and -5 to get 5.

$$12x+x^{2}+40=\left(x+5\right)^{2}+x-2\left(-5\right)+10$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+5\right)^{2}.

$$12x+x^{2}+40=x^{2}+10x+25+x-2\left(-5\right)+10$$

Combine 10x and x to get 11x.

$$12x+x^{2}+40=x^{2}+11x+25-2\left(-5\right)+10$$

Multiply -2 and -5 to get 10.

$$12x+x^{2}+40=x^{2}+11x+25+10+10$$

Add 25 and 10 to get 35.

$$12x+x^{2}+40=x^{2}+11x+35+10$$

Add 35 and 10 to get 45.

$$12x+x^{2}+40=x^{2}+11x+45$$

Subtract x^{2} from both sides.

$$12x+x^{2}+40-x^{2}=11x+45$$

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

$$12x+40=11x+45$$

Subtract 11x from both sides.

$$12x+40-11x=45$$

Combine 12x and -11x to get x.

$$x+40=45$$

Subtract 40 from both sides.

$$x=45-40$$

Subtract 40 from 45 to get 5.

$$x=5$$

The system is now solved.

$$x=5 y=-5$$