Solve {l}{x+y=10}{1000x-500y=8000}

$$\left\{ \begin{array} { l } { x + y = 10 } \\ { 1000 x - 500 y = 8000 } \end{array} \right.$$

Solve for x, y

$$x = \frac{26}{3} = 8\frac{2}{3} \approx 8.666666667$$

$$y = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

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.

$$x+y=10,1000x-500y=8000$$

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

$$x+y=10$$

Subtract y from both sides of the equation.

$$x=-y+10$$

Substitute -y+10 for x in the other equation, 1000x-500y=8000.

$$1000\left(-y+10\right)-500y=8000$$

Multiply 1000 times -y+10.

$$-1000y+10000-500y=8000$$

Add -1000y to -500y.

$$-1500y+10000=8000$$

Subtract 10000 from both sides of the equation.

$$-1500y=-2000$$

Divide both sides by -1500.

$$y=\frac{4}{3}$$

Substitute \frac{4}{3} for y in x=-y+10. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{3}+10$$

Add 10 to -\frac{4}{3}.

$$x=\frac{26}{3}$$

The system is now solved.

$$x=\frac{26}{3},y=\frac{4}{3}$$

Steps Using Matrices

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

$$x+y=10,1000x-500y=8000$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\8000\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1000&-500\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right))\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right))\left(\begin{matrix}10\\8000\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right))\left(\begin{matrix}10\\8000\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1000&-500\end{matrix}\right))\left(\begin{matrix}10\\8000\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{500}{-500-1000}&-\frac{1}{-500-1000}\\-\frac{1000}{-500-1000}&\frac{1}{-500-1000}\end{matrix}\right)\left(\begin{matrix}10\\8000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}&\frac{1}{1500}\\\frac{2}{3}&-\frac{1}{1500}\end{matrix}\right)\left(\begin{matrix}10\\8000\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3}\times 10+\frac{1}{1500}\times 8000\\\frac{2}{3}\times 10-\frac{1}{1500}\times 8000\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{26}{3}\\\frac{4}{3}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{26}{3},y=\frac{4}{3}$$

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.

$$x+y=10,1000x-500y=8000$$

To make x and 1000x equal, multiply all terms on each side of the first equation by 1000 and all terms on each side of the second by 1.

$$1000x+1000y=1000\times 10,1000x-500y=8000$$

Simplify.

$$1000x+1000y=10000,1000x-500y=8000$$

Subtract 1000x-500y=8000 from 1000x+1000y=10000 by subtracting like terms on each side of the equal sign.

$$1000x-1000x+1000y+500y=10000-8000$$

Add 1000x to -1000x. Terms 1000x and -1000x cancel out, leaving an equation with only one variable that can be solved.

$$1000y+500y=10000-8000$$

Add 1000y to 500y.

$$1500y=10000-8000$$

Add 10000 to -8000.

$$1500y=2000$$

Divide both sides by 1500.

$$y=\frac{4}{3}$$

Substitute \frac{4}{3} for y in 1000x-500y=8000. Because the resulting equation contains only one variable, you can solve for x directly.

$$1000x-500\times \frac{4}{3}=8000$$

Multiply -500 times \frac{4}{3}.

$$1000x-\frac{2000}{3}=8000$$

Add \frac{2000}{3} to both sides of the equation.

$$1000x=\frac{26000}{3}$$

Divide both sides by 1000.

$$x=\frac{26}{3}$$

The system is now solved.

$$x=\frac{26}{3},y=\frac{4}{3}$$