Solve {l}{13x=11y+12}{7x=6y+4}

$$\left\{ \begin{array} { l } { 13 x = 11 y + 12 } \\ { 7 x = 6 y + 4 } \end{array} \right.$$

Solve for x, y

$$x=28$$

$$y=32$$

Steps Using Substitution

Consider the first equation. Subtract 11y from both sides.

$$13x-11y=12$$

Consider the second equation. Subtract 6y from both sides.

$$7x-6y=4$$

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.

$$13x-11y=12,7x-6y=4$$

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

$$13x-11y=12$$

Add 11y to both sides of the equation.

$$13x=11y+12$$

Divide both sides by 13.

$$x=\frac{1}{13}\left(11y+12\right)$$

Multiply \frac{1}{13} times 11y+12.

$$x=\frac{11}{13}y+\frac{12}{13}$$

Substitute \frac{11y+12}{13} for x in the other equation, 7x-6y=4.

$$7\left(\frac{11}{13}y+\frac{12}{13}\right)-6y=4$$

Multiply 7 times \frac{11y+12}{13}.

$$\frac{77}{13}y+\frac{84}{13}-6y=4$$

Add \frac{77y}{13} to -6y.

$$-\frac{1}{13}y+\frac{84}{13}=4$$

Subtract \frac{84}{13} from both sides of the equation.

$$-\frac{1}{13}y=-\frac{32}{13}$$

Multiply both sides by -13.

$$y=32$$

Substitute 32 for y in x=\frac{11}{13}y+\frac{12}{13}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{11}{13}\times 32+\frac{12}{13}$$

Multiply \frac{11}{13} times 32.

$$x=\frac{352+12}{13}$$

Add \frac{12}{13} to \frac{352}{13} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=28$$

The system is now solved.

$$x=28,y=32$$

Steps Using Matrices

Consider the first equation. Subtract 11y from both sides.

$$13x-11y=12$$

Consider the second equation. Subtract 6y from both sides.

$$7x-6y=4$$

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

$$13x-11y=12,7x-6y=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}13&-11\\7&-6\end{matrix}\right).

$$inverse(\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right))\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right))\left(\begin{matrix}12\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right))\left(\begin{matrix}12\\4\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}13&-11\\7&-6\end{matrix}\right))\left(\begin{matrix}12\\4\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{13\left(-6\right)-\left(-11\times 7\right)}&-\frac{-11}{13\left(-6\right)-\left(-11\times 7\right)}\\-\frac{7}{13\left(-6\right)-\left(-11\times 7\right)}&\frac{13}{13\left(-6\right)-\left(-11\times 7\right)}\end{matrix}\right)\left(\begin{matrix}12\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6&-11\\7&-13\end{matrix}\right)\left(\begin{matrix}12\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\times 12-11\times 4\\7\times 12-13\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}28\\32\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=28,y=32$$

Steps Using Elimination

Consider the first equation. Subtract 11y from both sides.

$$13x-11y=12$$

Consider the second equation. Subtract 6y from both sides.

$$7x-6y=4$$

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.

$$13x-11y=12,7x-6y=4$$

To make 13x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by 13.

$$7\times 13x+7\left(-11\right)y=7\times 12,13\times 7x+13\left(-6\right)y=13\times 4$$

Simplify.

$$91x-77y=84,91x-78y=52$$

Subtract 91x-78y=52 from 91x-77y=84 by subtracting like terms on each side of the equal sign.

$$91x-91x-77y+78y=84-52$$

Add 91x to -91x. Terms 91x and -91x cancel out, leaving an equation with only one variable that can be solved.

$$-77y+78y=84-52$$

Add -77y to 78y.

$$y=84-52$$

Add 84 to -52.

$$y=32$$

Substitute 32 for y in 7x-6y=4. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x-6\times 32=4$$

Multiply -6 times 32.

$$7x-192=4$$

Add 192 to both sides of the equation.

$$7x=196$$

Divide both sides by 7.

$$x=28$$

The system is now solved.

$$x=28,y=32$$