Solve {c}{236-232=2*x-4*y}{232=20*(x+y)+4*y}

$$\left. \begin{array} { c } { 236 - 232 = 2 \cdot x - 4 \cdot y } \\ { 232 = 20 \cdot ( x + y ) + 4 \cdot y } \end{array} \right.$$

Solve for x, y

$$x=8$$

$$y=3$$

Steps Using Substitution

Consider the first equation. Subtract 232 from 236 to get 4.

$$4=2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2x-4y=4$$

Consider the second equation. Use the distributive property to multiply 20 by x+y.

$$232=20x+20y+4y$$

Combine 20y and 4y to get 24y.

$$232=20x+24y$$

Swap sides so that all variable terms are on the left hand side.

$$20x+24y=232$$

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.

$$2x-4y=4,20x+24y=232$$

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

$$2x-4y=4$$

Add 4y to both sides of the equation.

$$2x=4y+4$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(4y+4\right)$$

Multiply \frac{1}{2} times 4+4y.

$$x=2y+2$$

Substitute 2+2y for x in the other equation, 20x+24y=232.

$$20\left(2y+2\right)+24y=232$$

Multiply 20 times 2+2y.

$$40y+40+24y=232$$

Add 40y to 24y.

$$64y+40=232$$

Subtract 40 from both sides of the equation.

$$64y=192$$

Divide both sides by 64.

$$y=3$$

Substitute 3 for y in x=2y+2. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2\times 3+2$$

Multiply 2 times 3.

$$x=6+2$$

Add 2 to 6.

$$x=8$$

The system is now solved.

$$x=8,y=3$$

Steps Using Matrices

Consider the first equation. Subtract 232 from 236 to get 4.

$$4=2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2x-4y=4$$

Consider the second equation. Use the distributive property to multiply 20 by x+y.

$$232=20x+20y+4y$$

Combine 20y and 4y to get 24y.

$$232=20x+24y$$

Swap sides so that all variable terms are on the left hand side.

$$20x+24y=232$$

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

$$2x-4y=4,20x+24y=232$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-4\\20&24\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\\232\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-4\\20&24\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-4\\20&24\end{matrix}\right))\left(\begin{matrix}2&-4\\20&24\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-4\\20&24\end{matrix}\right))\left(\begin{matrix}4\\232\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-4\\20&24\end{matrix}\right))\left(\begin{matrix}4\\232\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-4\\20&24\end{matrix}\right))\left(\begin{matrix}4\\232\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{24}{2\times 24-\left(-4\times 20\right)}&-\frac{-4}{2\times 24-\left(-4\times 20\right)}\\-\frac{20}{2\times 24-\left(-4\times 20\right)}&\frac{2}{2\times 24-\left(-4\times 20\right)}\end{matrix}\right)\left(\begin{matrix}4\\232\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{16}&\frac{1}{32}\\-\frac{5}{32}&\frac{1}{64}\end{matrix}\right)\left(\begin{matrix}4\\232\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{16}\times 4+\frac{1}{32}\times 232\\-\frac{5}{32}\times 4+\frac{1}{64}\times 232\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=8,y=3$$

Steps Using Elimination

Consider the first equation. Subtract 232 from 236 to get 4.

$$4=2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2x-4y=4$$

Consider the second equation. Use the distributive property to multiply 20 by x+y.

$$232=20x+20y+4y$$

Combine 20y and 4y to get 24y.

$$232=20x+24y$$

Swap sides so that all variable terms are on the left hand side.

$$20x+24y=232$$

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.

$$2x-4y=4,20x+24y=232$$

To make 2x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 2.

$$20\times 2x+20\left(-4\right)y=20\times 4,2\times 20x+2\times 24y=2\times 232$$

Simplify.

$$40x-80y=80,40x+48y=464$$

Subtract 40x+48y=464 from 40x-80y=80 by subtracting like terms on each side of the equal sign.

$$40x-40x-80y-48y=80-464$$

Add 40x to -40x. Terms 40x and -40x cancel out, leaving an equation with only one variable that can be solved.

$$-80y-48y=80-464$$

Add -80y to -48y.

$$-128y=80-464$$

Add 80 to -464.

$$-128y=-384$$

Divide both sides by -128.

$$y=3$$

Substitute 3 for y in 20x+24y=232. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x+24\times 3=232$$

Multiply 24 times 3.

$$20x+72=232$$

Subtract 72 from both sides of the equation.

$$20x=160$$

Divide both sides by 20.

$$x=8$$

The system is now solved.

$$x=8,y=3$$