Solve {l}{15x+20y=8500}{10x+10y=5000}

$$\left\{ \begin{array} { l } { 15 x + 20 y = 8500 } \\ { 10 x + 10 y = 5000 } \end{array} \right.$$

Solve for x, y

$$x=300$$

$$y=200$$

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.

$$15x+20y=8500,10x+10y=5000$$

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

$$15x+20y=8500$$

Subtract 20y from both sides of the equation.

$$15x=-20y+8500$$

Divide both sides by 15.

$$x=\frac{1}{15}\left(-20y+8500\right)$$

Multiply \frac{1}{15} times -20y+8500.

$$x=-\frac{4}{3}y+\frac{1700}{3}$$

Substitute \frac{-4y+1700}{3} for x in the other equation, 10x+10y=5000.

$$10\left(-\frac{4}{3}y+\frac{1700}{3}\right)+10y=5000$$

Multiply 10 times \frac{-4y+1700}{3}.

$$-\frac{40}{3}y+\frac{17000}{3}+10y=5000$$

Add -\frac{40y}{3} to 10y.

$$-\frac{10}{3}y+\frac{17000}{3}=5000$$

Subtract \frac{17000}{3} from both sides of the equation.

$$-\frac{10}{3}y=-\frac{2000}{3}$$

Divide both sides of the equation by -\frac{10}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=200$$

Substitute 200 for y in x=-\frac{4}{3}y+\frac{1700}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{3}\times 200+\frac{1700}{3}$$

Multiply -\frac{4}{3} times 200.

$$x=\frac{-800+1700}{3}$$

Add \frac{1700}{3} to -\frac{800}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=300$$

The system is now solved.

$$x=300,y=200$$

Steps Using Matrices

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

$$15x+20y=8500,10x+10y=5000$$

Write the equations in matrix form.

$$\left(\begin{matrix}15&20\\10&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}15&20\\10&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}15&20\\10&10\end{matrix}\right))\left(\begin{matrix}15&20\\10&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&20\\10&10\end{matrix}\right))\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&20\\10&10\end{matrix}\right))\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}15&20\\10&10\end{matrix}\right))\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{15\times 10-20\times 10}&-\frac{20}{15\times 10-20\times 10}\\-\frac{10}{15\times 10-20\times 10}&\frac{15}{15\times 10-20\times 10}\end{matrix}\right)\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{2}{5}\\\frac{1}{5}&-\frac{3}{10}\end{matrix}\right)\left(\begin{matrix}8500\\5000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 8500+\frac{2}{5}\times 5000\\\frac{1}{5}\times 8500-\frac{3}{10}\times 5000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\200\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=300,y=200$$

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.

$$15x+20y=8500,10x+10y=5000$$

To make 15x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 15.

$$10\times 15x+10\times 20y=10\times 8500,15\times 10x+15\times 10y=15\times 5000$$

Simplify.

$$150x+200y=85000,150x+150y=75000$$

Subtract 150x+150y=75000 from 150x+200y=85000 by subtracting like terms on each side of the equal sign.

$$150x-150x+200y-150y=85000-75000$$

Add 150x to -150x. Terms 150x and -150x cancel out, leaving an equation with only one variable that can be solved.

$$200y-150y=85000-75000$$

Add 200y to -150y.

$$50y=85000-75000$$

Add 85000 to -75000.

$$50y=10000$$

Divide both sides by 50.

$$y=200$$

Substitute 200 for y in 10x+10y=5000. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+10\times 200=5000$$

Multiply 10 times 200.

$$10x+2000=5000$$

Subtract 2000 from both sides of the equation.

$$10x=3000$$

Divide both sides by 10.

$$x=300$$

The system is now solved.

$$x=300,y=200$$