Solve {l}{x-y=400}{80000/x=64000/y}

$$\left\{ \begin{array} { l } { x - y = 400 } \\ { \frac { 80000 } { x } = \frac { 64000 } { y } } \end{array} \right.$$

Solve for x, y

$$x=2000$$

$$y=1600$$

Steps Using Substitution

Consider the second equation. Variable x cannot be equal to 0 since division by zero is not defined. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by xy, the least common multiple of x,y.

$$y\times 80000=x\times 64000$$

Subtract x\times 64000 from both sides.

$$y\times 80000-x\times 64000=0$$

Multiply -1 and 64000 to get -64000.

$$y\times 80000-64000x=0$$

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.

$$x-y=400,-64000x+80000y=0$$

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

$$x-y=400$$

Add y to both sides of the equation.

$$x=y+400$$

Substitute y+400 for x in the other equation, -64000x+80000y=0.

$$-64000\left(y+400\right)+80000y=0$$

Multiply -64000 times y+400.

$$-64000y-25600000+80000y=0$$

Add -64000y to 80000y.

$$16000y-25600000=0$$

Add 25600000 to both sides of the equation.

$$16000y=25600000$$

Divide both sides by 16000.

$$y=1600$$

Substitute 1600 for y in x=y+400. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=1600+400$$

Add 400 to 1600.

$$x=2000$$

The system is now solved.

$$x=2000,y=1600$$

Steps Using Matrices

Consider the second equation. Variable x cannot be equal to 0 since division by zero is not defined. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by xy, the least common multiple of x,y.

$$y\times 80000=x\times 64000$$

Subtract x\times 64000 from both sides.

$$y\times 80000-x\times 64000=0$$

Multiply -1 and 64000 to get -64000.

$$y\times 80000-64000x=0$$

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

$$x-y=400,-64000x+80000y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}400\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right))\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right))\left(\begin{matrix}400\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right))\left(\begin{matrix}400\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\-64000&80000\end{matrix}\right))\left(\begin{matrix}400\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80000}{80000-\left(-\left(-64000\right)\right)}&-\frac{-1}{80000-\left(-\left(-64000\right)\right)}\\-\frac{-64000}{80000-\left(-\left(-64000\right)\right)}&\frac{1}{80000-\left(-\left(-64000\right)\right)}\end{matrix}\right)\left(\begin{matrix}400\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5&\frac{1}{16000}\\4&\frac{1}{16000}\end{matrix}\right)\left(\begin{matrix}400\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\times 400\\4\times 400\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\1600\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=2000,y=1600$$

Steps Using Elimination

Consider the second equation. Variable x cannot be equal to 0 since division by zero is not defined. Variable y cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by xy, the least common multiple of x,y.

$$y\times 80000=x\times 64000$$

Subtract x\times 64000 from both sides.

$$y\times 80000-x\times 64000=0$$

Multiply -1 and 64000 to get -64000.

$$y\times 80000-64000x=0$$

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.

$$x-y=400,-64000x+80000y=0$$

To make x and -64000x equal, multiply all terms on each side of the first equation by -64000 and all terms on each side of the second by 1.

$$-64000x-64000\left(-1\right)y=-64000\times 400,-64000x+80000y=0$$

Simplify.

$$-64000x+64000y=-25600000,-64000x+80000y=0$$

Subtract -64000x+80000y=0 from -64000x+64000y=-25600000 by subtracting like terms on each side of the equal sign.

$$-64000x+64000x+64000y-80000y=-25600000$$

Add -64000x to 64000x. Terms -64000x and 64000x cancel out, leaving an equation with only one variable that can be solved.

$$64000y-80000y=-25600000$$

Add 64000y to -80000y.

$$-16000y=-25600000$$

Divide both sides by -16000.

$$y=1600$$

Substitute 1600 for y in -64000x+80000y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-64000x+80000\times 1600=0$$

Multiply 80000 times 1600.

$$-64000x+128000000=0$$

Subtract 128000000 from both sides of the equation.

$$-64000x=-128000000$$

Divide both sides by -64000.

$$x=2000$$

The system is now solved.

$$x=2000,y=1600$$