Solve {l}{2x+2y-5/5=-13}{x-y+2/6=-10}

$$\left. \begin{array} { l } { 2 x + \frac { 2 y - 5 } { 5 } = - 13 } \\ { x - \frac { y + 2 } { 6 } = - 10 } \end{array} \right.$$

Solve for x, y

$$x=-8$$

$$y=10$$

Steps Using Substitution

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

$$10x+2y-5=-65$$

Add 5 to both sides.

$$10x+2y=-65+5$$

Add -65 and 5 to get -60.

$$10x+2y=-60$$

Consider the second equation. Multiply both sides of the equation by 6.

$$6x-\left(y+2\right)=-60$$

To find the opposite of y+2, find the opposite of each term.

$$6x-y-2=-60$$

Add 2 to both sides.

$$6x-y=-60+2$$

Add -60 and 2 to get -58.

$$6x-y=-58$$

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.

$$10x+2y=-60,6x-y=-58$$

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

$$10x+2y=-60$$

Subtract 2y from both sides of the equation.

$$10x=-2y-60$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-2y-60\right)$$

Multiply \frac{1}{10} times -2y-60.

$$x=-\frac{1}{5}y-6$$

Substitute -\frac{y}{5}-6 for x in the other equation, 6x-y=-58.

$$6\left(-\frac{1}{5}y-6\right)-y=-58$$

Multiply 6 times -\frac{y}{5}-6.

$$-\frac{6}{5}y-36-y=-58$$

Add -\frac{6y}{5} to -y.

$$-\frac{11}{5}y-36=-58$$

Add 36 to both sides of the equation.

$$-\frac{11}{5}y=-22$$

Divide both sides of the equation by -\frac{11}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=10$$

Substitute 10 for y in x=-\frac{1}{5}y-6. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{5}\times 10-6$$

Multiply -\frac{1}{5} times 10.

$$x=-2-6$$

Add -6 to -2.

$$x=-8$$

The system is now solved.

$$x=-8,y=10$$

Steps Using Matrices

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

$$10x+2y-5=-65$$

Add 5 to both sides.

$$10x+2y=-65+5$$

Add -65 and 5 to get -60.

$$10x+2y=-60$$

Consider the second equation. Multiply both sides of the equation by 6.

$$6x-\left(y+2\right)=-60$$

To find the opposite of y+2, find the opposite of each term.

$$6x-y-2=-60$$

Add 2 to both sides.

$$6x-y=-60+2$$

Add -60 and 2 to get -58.

$$6x-y=-58$$

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

$$10x+2y=-60,6x-y=-58$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&2\\6&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&2\\6&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&2\\6&-1\end{matrix}\right))\left(\begin{matrix}10&2\\6&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&2\\6&-1\end{matrix}\right))\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&2\\6&-1\end{matrix}\right))\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&2\\6&-1\end{matrix}\right))\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{10\left(-1\right)-2\times 6}&-\frac{2}{10\left(-1\right)-2\times 6}\\-\frac{6}{10\left(-1\right)-2\times 6}&\frac{10}{10\left(-1\right)-2\times 6}\end{matrix}\right)\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{22}&\frac{1}{11}\\\frac{3}{11}&-\frac{5}{11}\end{matrix}\right)\left(\begin{matrix}-60\\-58\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{22}\left(-60\right)+\frac{1}{11}\left(-58\right)\\\frac{3}{11}\left(-60\right)-\frac{5}{11}\left(-58\right)\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=-8,y=10$$

Steps Using Elimination

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

$$10x+2y-5=-65$$

Add 5 to both sides.

$$10x+2y=-65+5$$

Add -65 and 5 to get -60.

$$10x+2y=-60$$

Consider the second equation. Multiply both sides of the equation by 6.

$$6x-\left(y+2\right)=-60$$

To find the opposite of y+2, find the opposite of each term.

$$6x-y-2=-60$$

Add 2 to both sides.

$$6x-y=-60+2$$

Add -60 and 2 to get -58.

$$6x-y=-58$$

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.

$$10x+2y=-60,6x-y=-58$$

To make 10x and 6x equal, multiply all terms on each side of the first equation by 6 and all terms on each side of the second by 10.

$$6\times 10x+6\times 2y=6\left(-60\right),10\times 6x+10\left(-1\right)y=10\left(-58\right)$$

Simplify.

$$60x+12y=-360,60x-10y=-580$$

Subtract 60x-10y=-580 from 60x+12y=-360 by subtracting like terms on each side of the equal sign.

$$60x-60x+12y+10y=-360+580$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$12y+10y=-360+580$$

Add 12y to 10y.

$$22y=-360+580$$

Add -360 to 580.

$$22y=220$$

Divide both sides by 22.

$$y=10$$

Substitute 10 for y in 6x-y=-58. Because the resulting equation contains only one variable, you can solve for x directly.

$$6x-10=-58$$

Add 10 to both sides of the equation.

$$6x=-48$$

Divide both sides by 6.

$$x=-8$$

The system is now solved.

$$x=-8,y=10$$