Solve {l}{12y+20x=112}{12x+20y=144}

$$\left\{ \begin{array} { l } { 12 y + 20 x = 112 } \\ { 12 x + 20 y = 144 } \end{array} \right.$$

Solve for y, x

$$x=2$$

$$y=6$$

Steps Using Substitution

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.

$$12y+20x=112,20y+12x=144$$

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

$$12y+20x=112$$

Subtract 20x from both sides of the equation.

$$12y=-20x+112$$

Divide both sides by 12.

$$y=\frac{1}{12}\left(-20x+112\right)$$

Multiply \frac{1}{12} times -20x+112.

$$y=-\frac{5}{3}x+\frac{28}{3}$$

Substitute \frac{-5x+28}{3} for y in the other equation, 20y+12x=144.

$$20\left(-\frac{5}{3}x+\frac{28}{3}\right)+12x=144$$

Multiply 20 times \frac{-5x+28}{3}.

$$-\frac{100}{3}x+\frac{560}{3}+12x=144$$

Add -\frac{100x}{3} to 12x.

$$-\frac{64}{3}x+\frac{560}{3}=144$$

Subtract \frac{560}{3} from both sides of the equation.

$$-\frac{64}{3}x=-\frac{128}{3}$$

Divide both sides of the equation by -\frac{64}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$x=2$$

Substitute 2 for x in y=-\frac{5}{3}x+\frac{28}{3}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{5}{3}\times 2+\frac{28}{3}$$

Multiply -\frac{5}{3} times 2.

$$y=\frac{-10+28}{3}$$

Add \frac{28}{3} to -\frac{10}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=6$$

The system is now solved.

$$y=6,x=2$$

Steps Using Matrices

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

$$12y+20x=112,20y+12x=144$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&20\\20&12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}112\\144\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&20\\20&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&20\\20&12\end{matrix}\right))\left(\begin{matrix}12&20\\20&12\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}12&20\\20&12\end{matrix}\right))\left(\begin{matrix}112\\144\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}12&20\\20&12\end{matrix}\right))\left(\begin{matrix}112\\144\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}12&20\\20&12\end{matrix}\right))\left(\begin{matrix}112\\144\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{12}{12\times 12-20\times 20}&-\frac{20}{12\times 12-20\times 20}\\-\frac{20}{12\times 12-20\times 20}&\frac{12}{12\times 12-20\times 20}\end{matrix}\right)\left(\begin{matrix}112\\144\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{64}&\frac{5}{64}\\\frac{5}{64}&-\frac{3}{64}\end{matrix}\right)\left(\begin{matrix}112\\144\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{64}\times 112+\frac{5}{64}\times 144\\\frac{5}{64}\times 112-\frac{3}{64}\times 144\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}6\\2\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=6,x=2$$

Steps Using Elimination

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.

$$12y+20x=112,20y+12x=144$$

To make 12y and 20y equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 12.

$$20\times 12y+20\times 20x=20\times 112,12\times 20y+12\times 12x=12\times 144$$

Simplify.

$$240y+400x=2240,240y+144x=1728$$

Subtract 240y+144x=1728 from 240y+400x=2240 by subtracting like terms on each side of the equal sign.

$$240y-240y+400x-144x=2240-1728$$

Add 240y to -240y. Terms 240y and -240y cancel out, leaving an equation with only one variable that can be solved.

$$400x-144x=2240-1728$$

Add 400x to -144x.

$$256x=2240-1728$$

Add 2240 to -1728.

$$256x=512$$

Divide both sides by 256.

$$x=2$$

Substitute 2 for x in 20y+12x=144. Because the resulting equation contains only one variable, you can solve for y directly.

$$20y+12\times 2=144$$

Multiply 12 times 2.

$$20y+24=144$$

Subtract 24 from both sides of the equation.

$$20y=120$$

Divide both sides by 20.

$$y=6$$

The system is now solved.

$$y=6,x=2$$