Solve {l}{20(x-y)=600-500}{40(2y-x)=600-400}

$$\left\{ \begin{array} { l } { 20 ( x - y ) = 600 - 500 } \\ { 40 ( 2 y - x ) = 600 - 400 } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=10$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 20 by x-y.

$$20x-20y=600-500$$

Subtract 500 from 600 to get 100.

$$20x-20y=100$$

Consider the second equation. Use the distributive property to multiply 40 by 2y-x.

$$80y-40x=600-400$$

Subtract 400 from 600 to get 200.

$$80y-40x=200$$

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.

$$20x-20y=100,-40x+80y=200$$

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

$$20x-20y=100$$

Add 20y to both sides of the equation.

$$20x=20y+100$$

Divide both sides by 20.

$$x=\frac{1}{20}\left(20y+100\right)$$

Multiply \frac{1}{20} times 100+20y.

$$x=y+5$$

Substitute y+5 for x in the other equation, -40x+80y=200.

$$-40\left(y+5\right)+80y=200$$

Multiply -40 times y+5.

$$-40y-200+80y=200$$

Add -40y to 80y.

$$40y-200=200$$

Add 200 to both sides of the equation.

$$40y=400$$

Divide both sides by 40.

$$y=10$$

Substitute 10 for y in x=y+5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=10+5$$

Add 5 to 10.

$$x=15$$

The system is now solved.

$$x=15,y=10$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 20 by x-y.

$$20x-20y=600-500$$

Subtract 500 from 600 to get 100.

$$20x-20y=100$$

Consider the second equation. Use the distributive property to multiply 40 by 2y-x.

$$80y-40x=600-400$$

Subtract 400 from 600 to get 200.

$$80y-40x=200$$

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

$$20x-20y=100,-40x+80y=200$$

Write the equations in matrix form.

$$\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\200\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}20&-20\\-40&80\end{matrix}\right).

$$inverse(\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right))\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right))\left(\begin{matrix}100\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right))\left(\begin{matrix}100\\200\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}20&-20\\-40&80\end{matrix}\right))\left(\begin{matrix}100\\200\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{20\times 80-\left(-20\left(-40\right)\right)}&-\frac{-20}{20\times 80-\left(-20\left(-40\right)\right)}\\-\frac{-40}{20\times 80-\left(-20\left(-40\right)\right)}&\frac{20}{20\times 80-\left(-20\left(-40\right)\right)}\end{matrix}\right)\left(\begin{matrix}100\\200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}&\frac{1}{40}\\\frac{1}{20}&\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}100\\200\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}\times 100+\frac{1}{40}\times 200\\\frac{1}{20}\times 100+\frac{1}{40}\times 200\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=10$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 20 by x-y.

$$20x-20y=600-500$$

Subtract 500 from 600 to get 100.

$$20x-20y=100$$

Consider the second equation. Use the distributive property to multiply 40 by 2y-x.

$$80y-40x=600-400$$

Subtract 400 from 600 to get 200.

$$80y-40x=200$$

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.

$$20x-20y=100,-40x+80y=200$$

To make 20x and -40x equal, multiply all terms on each side of the first equation by -40 and all terms on each side of the second by 20.

$$-40\times 20x-40\left(-20\right)y=-40\times 100,20\left(-40\right)x+20\times 80y=20\times 200$$

Simplify.

$$-800x+800y=-4000,-800x+1600y=4000$$

Subtract -800x+1600y=4000 from -800x+800y=-4000 by subtracting like terms on each side of the equal sign.

$$-800x+800x+800y-1600y=-4000-4000$$

Add -800x to 800x. Terms -800x and 800x cancel out, leaving an equation with only one variable that can be solved.

$$800y-1600y=-4000-4000$$

Add 800y to -1600y.

$$-800y=-4000-4000$$

Add -4000 to -4000.

$$-800y=-8000$$

Divide both sides by -800.

$$y=10$$

Substitute 10 for y in -40x+80y=200. Because the resulting equation contains only one variable, you can solve for x directly.

$$-40x+80\times 10=200$$

Multiply 80 times 10.

$$-40x+800=200$$

Subtract 800 from both sides of the equation.

$$-40x=-600$$

Divide both sides by -40.

$$x=15$$

The system is now solved.

$$x=15,y=10$$