Solve {l}{x-y=500}{23x/20-9y/10=950}

$$\left\{ \begin{array} { l } { x - y = 500 } \\ { \frac { 23 x } { 20 } - \frac { 9 y } { 10 } = 950 } \end{array} \right.$$

Solve for x, y

$$x=2000$$

$$y=1500$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 20,10.

$$23x-2\times 9y=19000$$

Multiply -2 and 9 to get -18.

$$23x-18y=19000$$

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.

$$x-y=500,23x-18y=19000$$

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

$$x-y=500$$

Add y to both sides of the equation.

$$x=y+500$$

Substitute y+500 for x in the other equation, 23x-18y=19000.

$$23\left(y+500\right)-18y=19000$$

Multiply 23 times y+500.

$$23y+11500-18y=19000$$

Add 23y to -18y.

$$5y+11500=19000$$

Subtract 11500 from both sides of the equation.

$$5y=7500$$

Divide both sides by 5.

$$y=1500$$

Substitute 1500 for y in x=y+500. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=1500+500$$

Add 500 to 1500.

$$x=2000$$

The system is now solved.

$$x=2000,y=1500$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 20,10.

$$23x-2\times 9y=19000$$

Multiply -2 and 9 to get -18.

$$23x-18y=19000$$

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

$$x-y=500,23x-18y=19000$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\19000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\23&-18\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right))\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right))\left(\begin{matrix}500\\19000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right))\left(\begin{matrix}500\\19000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\23&-18\end{matrix}\right))\left(\begin{matrix}500\\19000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{18}{-18-\left(-23\right)}&-\frac{-1}{-18-\left(-23\right)}\\-\frac{23}{-18-\left(-23\right)}&\frac{1}{-18-\left(-23\right)}\end{matrix}\right)\left(\begin{matrix}500\\19000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{18}{5}&\frac{1}{5}\\-\frac{23}{5}&\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}500\\19000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{18}{5}\times 500+\frac{1}{5}\times 19000\\-\frac{23}{5}\times 500+\frac{1}{5}\times 19000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\1500\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2000,y=1500$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 20, the least common multiple of 20,10.

$$23x-2\times 9y=19000$$

Multiply -2 and 9 to get -18.

$$23x-18y=19000$$

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.

$$x-y=500,23x-18y=19000$$

To make x and 23x equal, multiply all terms on each side of the first equation by 23 and all terms on each side of the second by 1.

$$23x+23\left(-1\right)y=23\times 500,23x-18y=19000$$

Simplify.

$$23x-23y=11500,23x-18y=19000$$

Subtract 23x-18y=19000 from 23x-23y=11500 by subtracting like terms on each side of the equal sign.

$$23x-23x-23y+18y=11500-19000$$

Add 23x to -23x. Terms 23x and -23x cancel out, leaving an equation with only one variable that can be solved.

$$-23y+18y=11500-19000$$

Add -23y to 18y.

$$-5y=11500-19000$$

Add 11500 to -19000.

$$-5y=-7500$$

Divide both sides by -5.

$$y=1500$$

Substitute 1500 for y in 23x-18y=19000. Because the resulting equation contains only one variable, you can solve for x directly.

$$23x-18\times 1500=19000$$

Multiply -18 times 1500.

$$23x-27000=19000$$

Add 27000 to both sides of the equation.

$$23x=46000$$

Divide both sides by 23.

$$x=2000$$

The system is now solved.

$$x=2000,y=1500$$