Solve {l}{13x+11y=1392}{9(x+y)=1080}

$$\left\{ \begin{array} { l } { 13 x + 11 y = 1392 } \\ { 9 ( x + y ) = 1080 } \end{array} \right.$$

Solve for x, y

$$x=36$$

$$y=84$$

Steps Using Substitution

Consider the second equation. Divide both sides by 9.

$$x+y=\frac{1080}{9}$$

Divide 1080 by 9 to get 120.

$$x+y=120$$

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.

$$13x+11y=1392,x+y=120$$

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

$$13x+11y=1392$$

Subtract 11y from both sides of the equation.

$$13x=-11y+1392$$

Divide both sides by 13.

$$x=\frac{1}{13}\left(-11y+1392\right)$$

Multiply \frac{1}{13} times -11y+1392.

$$x=-\frac{11}{13}y+\frac{1392}{13}$$

Substitute \frac{-11y+1392}{13} for x in the other equation, x+y=120.

$$-\frac{11}{13}y+\frac{1392}{13}+y=120$$

Add -\frac{11y}{13} to y.

$$\frac{2}{13}y+\frac{1392}{13}=120$$

Subtract \frac{1392}{13} from both sides of the equation.

$$\frac{2}{13}y=\frac{168}{13}$$

Divide both sides of the equation by \frac{2}{13}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=84$$

Substitute 84 for y in x=-\frac{11}{13}y+\frac{1392}{13}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{11}{13}\times 84+\frac{1392}{13}$$

Multiply -\frac{11}{13} times 84.

$$x=\frac{-924+1392}{13}$$

Add \frac{1392}{13} to -\frac{924}{13} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=36$$

The system is now solved.

$$x=36,y=84$$

Steps Using Matrices

Consider the second equation. Divide both sides by 9.

$$x+y=\frac{1080}{9}$$

Divide 1080 by 9 to get 120.

$$x+y=120$$

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

$$13x+11y=1392,x+y=120$$

Write the equations in matrix form.

$$\left(\begin{matrix}13&11\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1392\\120\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}13&11\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}13&11\\1&1\end{matrix}\right))\left(\begin{matrix}13&11\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&11\\1&1\end{matrix}\right))\left(\begin{matrix}1392\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&11\\1&1\end{matrix}\right))\left(\begin{matrix}1392\\120\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&11\\1&1\end{matrix}\right))\left(\begin{matrix}1392\\120\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{13-11}&-\frac{11}{13-11}\\-\frac{1}{13-11}&\frac{13}{13-11}\end{matrix}\right)\left(\begin{matrix}1392\\120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&-\frac{11}{2}\\-\frac{1}{2}&\frac{13}{2}\end{matrix}\right)\left(\begin{matrix}1392\\120\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 1392-\frac{11}{2}\times 120\\-\frac{1}{2}\times 1392+\frac{13}{2}\times 120\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\84\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=36,y=84$$

Steps Using Elimination

Consider the second equation. Divide both sides by 9.

$$x+y=\frac{1080}{9}$$

Divide 1080 by 9 to get 120.

$$x+y=120$$

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.

$$13x+11y=1392,x+y=120$$

To make 13x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 13.

$$13x+11y=1392,13x+13y=13\times 120$$

Simplify.

$$13x+11y=1392,13x+13y=1560$$

Subtract 13x+13y=1560 from 13x+11y=1392 by subtracting like terms on each side of the equal sign.

$$13x-13x+11y-13y=1392-1560$$

Add 13x to -13x. Terms 13x and -13x cancel out, leaving an equation with only one variable that can be solved.

$$11y-13y=1392-1560$$

Add 11y to -13y.

$$-2y=1392-1560$$

Add 1392 to -1560.

$$-2y=-168$$

Divide both sides by -2.

$$y=84$$

Substitute 84 for y in x+y=120. Because the resulting equation contains only one variable, you can solve for x directly.

$$x+84=120$$

Subtract 84 from both sides of the equation.

$$x=36$$

The system is now solved.

$$x=36,y=84$$