Solve {l}{-36x-48x+12y=0}{2x-4x-16x+12y=0}

$$\left\{ \begin{array} { l } { - 36 x - 48 x + 12 y = 0 } \\ { 2 x - 4 x - 16 x + 12 y = 0 } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Combine -36x and -48x to get -84x.

$$-84x+12y=0$$

Consider the second equation. Combine 2x and -4x to get -2x.

$$-2x-16x+12y=0$$

Combine -2x and -16x to get -18x.

$$-18x+12y=0$$

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.

$$-84x+12y=0,-18x+12y=0$$

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

$$-84x+12y=0$$

Subtract 12y from both sides of the equation.

$$-84x=-12y$$

Divide both sides by -84.

$$x=-\frac{1}{84}\left(-12\right)y$$

Multiply -\frac{1}{84} times -12y.

$$x=\frac{1}{7}y$$

Substitute \frac{y}{7} for x in the other equation, -18x+12y=0.

$$-18\times \frac{1}{7}y+12y=0$$

Multiply -18 times \frac{y}{7}.

$$-\frac{18}{7}y+12y=0$$

Add -\frac{18y}{7} to 12y.

$$\frac{66}{7}y=0$$

Divide both sides of the equation by \frac{66}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=0$$

Substitute 0 for y in x=\frac{1}{7}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. Combine -36x and -48x to get -84x.

$$-84x+12y=0$$

Consider the second equation. Combine 2x and -4x to get -2x.

$$-2x-16x+12y=0$$

Combine -2x and -16x to get -18x.

$$-18x+12y=0$$

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

$$-84x+12y=0,-18x+12y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-84&12\\-18&12\end{matrix}\right).

$$inverse(\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right))\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-84&12\\-18&12\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{12}{-84\times 12-12\left(-18\right)}&-\frac{12}{-84\times 12-12\left(-18\right)}\\-\frac{-18}{-84\times 12-12\left(-18\right)}&-\frac{84}{-84\times 12-12\left(-18\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{66}&\frac{1}{66}\\-\frac{1}{44}&\frac{7}{66}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

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. Combine -36x and -48x to get -84x.

$$-84x+12y=0$$

Consider the second equation. Combine 2x and -4x to get -2x.

$$-2x-16x+12y=0$$

Combine -2x and -16x to get -18x.

$$-18x+12y=0$$

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.

$$-84x+12y=0,-18x+12y=0$$

Subtract -18x+12y=0 from -84x+12y=0 by subtracting like terms on each side of the equal sign.

$$-84x+18x+12y-12y=0$$

Add 12y to -12y. Terms 12y and -12y cancel out, leaving an equation with only one variable that can be solved.

$$-84x+18x=0$$

Add -84x to 18x.

$$-66x=0$$

Divide both sides by -66.

$$x=0$$

Substitute 0 for x in -18x+12y=0. Because the resulting equation contains only one variable, you can solve for y directly.

$$12y=0$$

Divide both sides by 12.

$$y=0$$

The system is now solved.

$$x=0,y=0$$