Solve {l}{8(x+5)+7(y-2)=-5}{2(x+6)-4-y/5=5}

$$\left. \begin{array} { l } { 8 ( x + 5 ) + 7 ( y - 2 ) = - 5 } \\ { 2 ( x + 6 ) - \frac { 4 - y } { 5 } = 5 } \end{array} \right.$$

Solve for x, y

$$x=-3$$

$$y=-1$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 8 by x+5.

$$8x+40+7\left(y-2\right)=-5$$

Use the distributive property to multiply 7 by y-2.

$$8x+40+7y-14=-5$$

Subtract 14 from 40 to get 26.

$$8x+26+7y=-5$$

Subtract 26 from both sides.

$$8x+7y=-5-26$$

Subtract 26 from -5 to get -31.

$$8x+7y=-31$$

Consider the second equation. Multiply both sides of the equation by 5.

$$10\left(x+6\right)-\left(4-y\right)=25$$

Use the distributive property to multiply 10 by x+6.

$$10x+60-\left(4-y\right)=25$$

To find the opposite of 4-y, find the opposite of each term.

$$10x+60-4+y=25$$

Subtract 4 from 60 to get 56.

$$10x+56+y=25$$

Subtract 56 from both sides.

$$10x+y=25-56$$

Subtract 56 from 25 to get -31.

$$10x+y=-31$$

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.

$$8x+7y=-31,10x+y=-31$$

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

$$8x+7y=-31$$

Subtract 7y from both sides of the equation.

$$8x=-7y-31$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-7y-31\right)$$

Multiply \frac{1}{8} times -7y-31.

$$x=-\frac{7}{8}y-\frac{31}{8}$$

Substitute \frac{-7y-31}{8} for x in the other equation, 10x+y=-31.

$$10\left(-\frac{7}{8}y-\frac{31}{8}\right)+y=-31$$

Multiply 10 times \frac{-7y-31}{8}.

$$-\frac{35}{4}y-\frac{155}{4}+y=-31$$

Add -\frac{35y}{4} to y.

$$-\frac{31}{4}y-\frac{155}{4}=-31$$

Add \frac{155}{4} to both sides of the equation.

$$-\frac{31}{4}y=\frac{31}{4}$$

Divide both sides of the equation by -\frac{31}{4}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-1$$

Substitute -1 for y in x=-\frac{7}{8}y-\frac{31}{8}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{7}{8}\left(-1\right)-\frac{31}{8}$$

Multiply -\frac{7}{8} times -1.

$$x=\frac{7-31}{8}$$

Add -\frac{31}{8} to \frac{7}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-3$$

The system is now solved.

$$x=-3,y=-1$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 8 by x+5.

$$8x+40+7\left(y-2\right)=-5$$

Use the distributive property to multiply 7 by y-2.

$$8x+40+7y-14=-5$$

Subtract 14 from 40 to get 26.

$$8x+26+7y=-5$$

Subtract 26 from both sides.

$$8x+7y=-5-26$$

Subtract 26 from -5 to get -31.

$$8x+7y=-31$$

Consider the second equation. Multiply both sides of the equation by 5.

$$10\left(x+6\right)-\left(4-y\right)=25$$

Use the distributive property to multiply 10 by x+6.

$$10x+60-\left(4-y\right)=25$$

To find the opposite of 4-y, find the opposite of each term.

$$10x+60-4+y=25$$

Subtract 4 from 60 to get 56.

$$10x+56+y=25$$

Subtract 56 from both sides.

$$10x+y=25-56$$

Subtract 56 from 25 to get -31.

$$10x+y=-31$$

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

$$8x+7y=-31,10x+y=-31$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&7\\10&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&7\\10&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&7\\10&1\end{matrix}\right))\left(\begin{matrix}8&7\\10&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&7\\10&1\end{matrix}\right))\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&7\\10&1\end{matrix}\right))\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&7\\10&1\end{matrix}\right))\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{8-7\times 10}&-\frac{7}{8-7\times 10}\\-\frac{10}{8-7\times 10}&\frac{8}{8-7\times 10}\end{matrix}\right)\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{62}&\frac{7}{62}\\\frac{5}{31}&-\frac{4}{31}\end{matrix}\right)\left(\begin{matrix}-31\\-31\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{62}\left(-31\right)+\frac{7}{62}\left(-31\right)\\\frac{5}{31}\left(-31\right)-\frac{4}{31}\left(-31\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\\-1\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-3,y=-1$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 8 by x+5.

$$8x+40+7\left(y-2\right)=-5$$

Use the distributive property to multiply 7 by y-2.

$$8x+40+7y-14=-5$$

Subtract 14 from 40 to get 26.

$$8x+26+7y=-5$$

Subtract 26 from both sides.

$$8x+7y=-5-26$$

Subtract 26 from -5 to get -31.

$$8x+7y=-31$$

Consider the second equation. Multiply both sides of the equation by 5.

$$10\left(x+6\right)-\left(4-y\right)=25$$

Use the distributive property to multiply 10 by x+6.

$$10x+60-\left(4-y\right)=25$$

To find the opposite of 4-y, find the opposite of each term.

$$10x+60-4+y=25$$

Subtract 4 from 60 to get 56.

$$10x+56+y=25$$

Subtract 56 from both sides.

$$10x+y=25-56$$

Subtract 56 from 25 to get -31.

$$10x+y=-31$$

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.

$$8x+7y=-31,10x+y=-31$$

To make 8x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 8.

$$10\times 8x+10\times 7y=10\left(-31\right),8\times 10x+8y=8\left(-31\right)$$

Simplify.

$$80x+70y=-310,80x+8y=-248$$

Subtract 80x+8y=-248 from 80x+70y=-310 by subtracting like terms on each side of the equal sign.

$$80x-80x+70y-8y=-310+248$$

Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.

$$70y-8y=-310+248$$

Add 70y to -8y.

$$62y=-310+248$$

Add -310 to 248.

$$62y=-62$$

Divide both sides by 62.

$$y=-1$$

Substitute -1 for y in 10x+y=-31. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x-1=-31$$

Add 1 to both sides of the equation.

$$10x=-30$$

Divide both sides by 10.

$$x=-3$$

The system is now solved.

$$x=-3,y=-1$$