Solve {l}{2x+8y=3x}{80y=5x}

$$\left\{ \begin{array} { l } { 2 x + 8 y = 3 x } \\ { 80 y = 5 x } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract 3x from both sides.

$$2x+8y-3x=0$$

Combine 2x and -3x to get -x.

$$-x+8y=0$$

Consider the second equation. Subtract 5x from both sides.

$$80y-5x=0$$

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.

$$-x+8y=0,-5x+80y=0$$

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

$$-x+8y=0$$

Subtract 8y from both sides of the equation.

$$-x=-8y$$

Divide both sides by -1.

$$x=-\left(-8\right)y$$

Multiply -1 times -8y.

$$x=8y$$

Substitute 8y for x in the other equation, -5x+80y=0.

$$-5\times 8y+80y=0$$

Multiply -5 times 8y.

$$-40y+80y=0$$

Add -40y to 80y.

$$40y=0$$

Divide both sides by 40.

$$y=0$$

Substitute 0 for y in x=8y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,y=0$$

Steps Using Matrices

Consider the first equation. Subtract 3x from both sides.

$$2x+8y-3x=0$$

Combine 2x and -3x to get -x.

$$-x+8y=0$$

Consider the second equation. Subtract 5x from both sides.

$$80y-5x=0$$

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

$$-x+8y=0,-5x+80y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-1&8\\-5&80\end{matrix}\right).

$$inverse(\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right))\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-1&8\\-5&80\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{-80-8\left(-5\right)}&-\frac{8}{-80-8\left(-5\right)}\\-\frac{-5}{-80-8\left(-5\right)}&-\frac{1}{-80-8\left(-5\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&\frac{1}{5}\\-\frac{1}{8}&\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=0,y=0$$

Steps Using Elimination

Consider the first equation. Subtract 3x from both sides.

$$2x+8y-3x=0$$

Combine 2x and -3x to get -x.

$$-x+8y=0$$

Consider the second equation. Subtract 5x from both sides.

$$80y-5x=0$$

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.

$$-x+8y=0,-5x+80y=0$$

To make -x and -5x equal, multiply all terms on each side of the first equation by -5 and all terms on each side of the second by -1.

$$-5\left(-1\right)x-5\times 8y=0,-\left(-5\right)x-80y=0$$

Simplify.

$$5x-40y=0,5x-80y=0$$

Subtract 5x-80y=0 from 5x-40y=0 by subtracting like terms on each side of the equal sign.

$$5x-5x-40y+80y=0$$

Add 5x to -5x. Terms 5x and -5x cancel out, leaving an equation with only one variable that can be solved.

$$-40y+80y=0$$

Add -40y to 80y.

$$40y=0$$

Divide both sides by 40.

$$y=0$$

Substitute 0 for y in -5x+80y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-5x=0$$

Divide both sides by -5.

$$x=0$$

The system is now solved.

$$x=0,y=0$$