Solve {l}{(x-2)^2=(x+5)^2+8y}{(y+1)^2=5x+y^2+6}

$$\left\{ \begin{array} { l } { ( x - 2 ) ^ { 2 } = ( x + 5 ) ^ { 2 } + 8 y } \\ { ( y + 1 ) ^ { 2 } = 5 x + y ^ { 2 } + 6 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{41}{34} = -1\frac{7}{34} \approx -1.205882353$$

$$y=-\frac{35}{68}\approx -0.514705882$$

Steps Using Substitution

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

$$x^{2}-4x+4=\left(x+5\right)^{2}+8y$$

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

$$x^{2}-4x+4=x^{2}+10x+25+8y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4-x^{2}=10x+25+8y$$

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

$$-4x+4=10x+25+8y$$

Subtract 10x from both sides.

$$-4x+4-10x=25+8y$$

Combine -4x and -10x to get -14x.

$$-14x+4=25+8y$$

Subtract 8y from both sides.

$$-14x+4-8y=25$$

Subtract 4 from both sides.

$$-14x-8y=25-4$$

Subtract 4 from 25 to get 21.

$$-14x-8y=21$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=5x+y^{2}+6$$

Subtract 5x from both sides.

$$y^{2}+2y+1-5x=y^{2}+6$$

Subtract y^{2} from both sides.

$$y^{2}+2y+1-5x-y^{2}=6$$

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

$$2y+1-5x=6$$

Subtract 1 from both sides.

$$2y-5x=6-1$$

Subtract 1 from 6 to get 5.

$$2y-5x=5$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$-14x-8y=21,-5x+2y=5$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$-14x-8y=21$$

Add 8y to both sides of the equation.

$$-14x=8y+21$$

Divide both sides by -14.

$$x=-\frac{1}{14}\left(8y+21\right)$$

Multiply -\frac{1}{14} times 8y+21.

$$x=-\frac{4}{7}y-\frac{3}{2}$$

Substitute -\frac{4y}{7}-\frac{3}{2} for x in the other equation, -5x+2y=5.

$$-5\left(-\frac{4}{7}y-\frac{3}{2}\right)+2y=5$$

Multiply -5 times -\frac{4y}{7}-\frac{3}{2}.

$$\frac{20}{7}y+\frac{15}{2}+2y=5$$

Add \frac{20y}{7} to 2y.

$$\frac{34}{7}y+\frac{15}{2}=5$$

Subtract \frac{15}{2} from both sides of the equation.

$$\frac{34}{7}y=-\frac{5}{2}$$

Divide both sides of the equation by \frac{34}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-\frac{35}{68}$$

Substitute -\frac{35}{68} for y in x=-\frac{4}{7}y-\frac{3}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{7}\left(-\frac{35}{68}\right)-\frac{3}{2}$$

Multiply -\frac{4}{7} times -\frac{35}{68} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=\frac{5}{17}-\frac{3}{2}$$

Add -\frac{3}{2} to \frac{5}{17} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{41}{34}$$

The system is now solved.

$$x=-\frac{41}{34},y=-\frac{35}{68}$$

Steps Using Matrices

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

$$x^{2}-4x+4=\left(x+5\right)^{2}+8y$$

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

$$x^{2}-4x+4=x^{2}+10x+25+8y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4-x^{2}=10x+25+8y$$

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

$$-4x+4=10x+25+8y$$

Subtract 10x from both sides.

$$-4x+4-10x=25+8y$$

Combine -4x and -10x to get -14x.

$$-14x+4=25+8y$$

Subtract 8y from both sides.

$$-14x+4-8y=25$$

Subtract 4 from both sides.

$$-14x-8y=25-4$$

Subtract 4 from 25 to get 21.

$$-14x-8y=21$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=5x+y^{2}+6$$

Subtract 5x from both sides.

$$y^{2}+2y+1-5x=y^{2}+6$$

Subtract y^{2} from both sides.

$$y^{2}+2y+1-5x-y^{2}=6$$

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

$$2y+1-5x=6$$

Subtract 1 from both sides.

$$2y-5x=6-1$$

Subtract 1 from 6 to get 5.

$$2y-5x=5$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$-14x-8y=21,-5x+2y=5$$

Write the equations in matrix form.

$$\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}21\\5\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right))\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right))\left(\begin{matrix}21\\5\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right))\left(\begin{matrix}21\\5\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&-8\\-5&2\end{matrix}\right))\left(\begin{matrix}21\\5\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{-14\times 2-\left(-8\left(-5\right)\right)}&-\frac{-8}{-14\times 2-\left(-8\left(-5\right)\right)}\\-\frac{-5}{-14\times 2-\left(-8\left(-5\right)\right)}&-\frac{14}{-14\times 2-\left(-8\left(-5\right)\right)}\end{matrix}\right)\left(\begin{matrix}21\\5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{34}&-\frac{2}{17}\\-\frac{5}{68}&\frac{7}{34}\end{matrix}\right)\left(\begin{matrix}21\\5\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{34}\times 21-\frac{2}{17}\times 5\\-\frac{5}{68}\times 21+\frac{7}{34}\times 5\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{41}{34}\\-\frac{35}{68}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{41}{34},y=-\frac{35}{68}$$

Steps Using Elimination

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

$$x^{2}-4x+4=\left(x+5\right)^{2}+8y$$

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

$$x^{2}-4x+4=x^{2}+10x+25+8y$$

Subtract x^{2} from both sides.

$$x^{2}-4x+4-x^{2}=10x+25+8y$$

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

$$-4x+4=10x+25+8y$$

Subtract 10x from both sides.

$$-4x+4-10x=25+8y$$

Combine -4x and -10x to get -14x.

$$-14x+4=25+8y$$

Subtract 8y from both sides.

$$-14x+4-8y=25$$

Subtract 4 from both sides.

$$-14x-8y=25-4$$

Subtract 4 from 25 to get 21.

$$-14x-8y=21$$

Consider the second equation. Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(y+1\right)^{2}.

$$y^{2}+2y+1=5x+y^{2}+6$$

Subtract 5x from both sides.

$$y^{2}+2y+1-5x=y^{2}+6$$

Subtract y^{2} from both sides.

$$y^{2}+2y+1-5x-y^{2}=6$$

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

$$2y+1-5x=6$$

Subtract 1 from both sides.

$$2y-5x=6-1$$

Subtract 1 from 6 to get 5.

$$2y-5x=5$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$-14x-8y=21,-5x+2y=5$$

To make -14x and -5x equal, multiply all terms on each side of the first equation by -5 and all terms on each side of the second by -14.

$$-5\left(-14\right)x-5\left(-8\right)y=-5\times 21,-14\left(-5\right)x-14\times 2y=-14\times 5$$

Simplify.

$$70x+40y=-105,70x-28y=-70$$

Subtract 70x-28y=-70 from 70x+40y=-105 by subtracting like terms on each side of the equal sign.

$$70x-70x+40y+28y=-105+70$$

Add 70x to -70x. Terms 70x and -70x cancel out, leaving an equation with only one variable that can be solved.

$$40y+28y=-105+70$$

Add 40y to 28y.

$$68y=-105+70$$

Add -105 to 70.

$$68y=-35$$

Divide both sides by 68.

$$y=-\frac{35}{68}$$

Substitute -\frac{35}{68} for y in -5x+2y=5. Because the resulting equation contains only one variable, you can solve for x directly.

$$-5x+2\left(-\frac{35}{68}\right)=5$$

Multiply 2 times -\frac{35}{68}.

$$-5x-\frac{35}{34}=5$$

Add \frac{35}{34} to both sides of the equation.

$$-5x=\frac{205}{34}$$

Divide both sides by -5.

$$x=-\frac{41}{34}$$

The system is now solved.

$$x=-\frac{41}{34},y=-\frac{35}{68}$$