Solve {l}{x+y=100}{80x=320y}

$$\left. \begin{array} { l } { x + y = 100 } \\ { 80 x = 320 y } \end{array} \right.$$

Solve for x, y

$$x=80$$

$$y=20$$

Steps Using Substitution

Consider the second equation. Subtract 320y from both sides.

$$80x-320y=0$$

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.

$$x+y=100,80x-320y=0$$

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

$$x+y=100$$

Subtract y from both sides of the equation.

$$x=-y+100$$

Substitute -y+100 for x in the other equation, 80x-320y=0.

$$80\left(-y+100\right)-320y=0$$

Multiply 80 times -y+100.

$$-80y+8000-320y=0$$

Add -80y to -320y.

$$-400y+8000=0$$

Subtract 8000 from both sides of the equation.

$$-400y=-8000$$

Divide both sides by -400.

$$y=20$$

Substitute 20 for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-20+100$$

Add 100 to -20.

$$x=80$$

The system is now solved.

$$x=80,y=20$$

Steps Using Matrices

Consider the second equation. Subtract 320y from both sides.

$$80x-320y=0$$

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

$$x+y=100,80x-320y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\80&-320\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\80&-320\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\80&-320\end{matrix}\right))\left(\begin{matrix}1&1\\80&-320\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\80&-320\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\80&-320\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\80&-320\end{matrix}\right))\left(\begin{matrix}100\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{320}{-320-80}&-\frac{1}{-320-80}\\-\frac{80}{-320-80}&\frac{1}{-320-80}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5}&\frac{1}{400}\\\frac{1}{5}&-\frac{1}{400}\end{matrix}\right)\left(\begin{matrix}100\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{5}\times 100\\\frac{1}{5}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=80,y=20$$

Steps Using Elimination

Consider the second equation. Subtract 320y from both sides.

$$80x-320y=0$$

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.

$$x+y=100,80x-320y=0$$

To make x and 80x equal, multiply all terms on each side of the first equation by 80 and all terms on each side of the second by 1.

$$80x+80y=80\times 100,80x-320y=0$$

Simplify.

$$80x+80y=8000,80x-320y=0$$

Subtract 80x-320y=0 from 80x+80y=8000 by subtracting like terms on each side of the equal sign.

$$80x-80x+80y+320y=8000$$

Add 80x to -80x. Terms 80x and -80x cancel out, leaving an equation with only one variable that can be solved.

$$80y+320y=8000$$

Add 80y to 320y.

$$400y=8000$$

Divide both sides by 400.

$$y=20$$

Substitute 20 for y in 80x-320y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$80x-320\times 20=0$$

Multiply -320 times 20.

$$80x-6400=0$$

Add 6400 to both sides of the equation.

$$80x=6400$$

Divide both sides by 80.

$$x=80$$

The system is now solved.

$$x=80,y=20$$