Solve {l}{1x+7y=8x}{9y+7x=15y}

$$\left\{ \begin{array} { l } { 1 x + 7 y = 8 x } \\ { 9 y + 7 x = 15 y } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract 8x from both sides.

$$1x+7y-8x=0$$

Combine 1x and -8x to get -7x.

$$-7x+7y=0$$

Consider the second equation. Subtract 15y from both sides.

$$9y+7x-15y=0$$

Combine 9y and -15y to get -6y.

$$-6y+7x=0$$

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.

$$-7x+7y=0,7x-6y=0$$

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

$$-7x+7y=0$$

Subtract 7y from both sides of the equation.

$$-7x=-7y$$

Divide both sides by -7.

$$x=-\frac{1}{7}\left(-7\right)y$$

Multiply -\frac{1}{7} times -7y.

$$x=y$$

Substitute y for x in the other equation, 7x-6y=0.

$$7y-6y=0$$

Add 7y to -6y.

$$y=0$$

Substitute 0 for y in x=y. 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 8x from both sides.

$$1x+7y-8x=0$$

Combine 1x and -8x to get -7x.

$$-7x+7y=0$$

Consider the second equation. Subtract 15y from both sides.

$$9y+7x-15y=0$$

Combine 9y and -15y to get -6y.

$$-6y+7x=0$$

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

$$-7x+7y=0,7x-6y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-7&7\\7&-6\end{matrix}\right).

$$inverse(\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right))\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-7&7\\7&-6\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{6}{-7\left(-6\right)-7\times 7}&-\frac{7}{-7\left(-6\right)-7\times 7}\\-\frac{7}{-7\left(-6\right)-7\times 7}&-\frac{7}{-7\left(-6\right)-7\times 7}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{6}{7}&1\\1&1\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

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 8x from both sides.

$$1x+7y-8x=0$$

Combine 1x and -8x to get -7x.

$$-7x+7y=0$$

Consider the second equation. Subtract 15y from both sides.

$$9y+7x-15y=0$$

Combine 9y and -15y to get -6y.

$$-6y+7x=0$$

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.

$$-7x+7y=0,7x-6y=0$$

To make -7x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by -7.

$$7\left(-7\right)x+7\times 7y=0,-7\times 7x-7\left(-6\right)y=0$$

Simplify.

$$-49x+49y=0,-49x+42y=0$$

Subtract -49x+42y=0 from -49x+49y=0 by subtracting like terms on each side of the equal sign.

$$-49x+49x+49y-42y=0$$

Add -49x to 49x. Terms -49x and 49x cancel out, leaving an equation with only one variable that can be solved.

$$49y-42y=0$$

Add 49y to -42y.

$$7y=0$$

Divide both sides by 7.

$$y=0$$

Substitute 0 for y in 7x-6y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x=0$$

Divide both sides by 7.

$$x=0$$

The system is now solved.

$$x=0,y=0$$