Solve {l}{x=3y(-10)}{y=x/3(-10)}

$$\left. \begin{array} { l } { x = 3 y ( - 10 ) } \\ { y = \frac { x } { 3 } ( - 10 ) } \end{array} \right.$$

Solve for x, y

$$x=0$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Multiply 3 and -10 to get -30.

$$x=-30y$$

Substitute -30y for x in the other equation, 10x+3y=0.

$$10\left(-30\right)y+3y=0$$

Multiply 10 times -30y.

$$-300y+3y=0$$

Add -300y to 3y.

$$-297y=0$$

Divide both sides by -297.

$$y=0$$

Substitute 0 for y in x=-30y. 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. Multiply 3 and -10 to get -30.

$$x=-30y$$

Add 30y to both sides.

$$x+30y=0$$

Consider the second equation. Express \frac{x}{3}\left(-10\right) as a single fraction.

$$y=\frac{-x\times 10}{3}$$

Multiply -1 and 10 to get -10.

$$y=\frac{-10x}{3}$$

Subtract \frac{-10x}{3} from both sides.

$$y-\frac{-10x}{3}=0$$

Multiply both sides of the equation by 3.

$$3y+10x=0$$

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

$$x+30y=0,10x+3y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&30\\10&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&30\\10&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&30\\10&3\end{matrix}\right))\left(\begin{matrix}1&30\\10&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&30\\10&3\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&30\\10&3\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&30\\10&3\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3-30\times 10}&-\frac{30}{3-30\times 10}\\-\frac{10}{3-30\times 10}&\frac{1}{3-30\times 10}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{99}&\frac{10}{99}\\\frac{10}{297}&-\frac{1}{297}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

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. Multiply 3 and -10 to get -30.

$$x=-30y$$

Add 30y to both sides.

$$x+30y=0$$

Consider the second equation. Express \frac{x}{3}\left(-10\right) as a single fraction.

$$y=\frac{-x\times 10}{3}$$

Multiply -1 and 10 to get -10.

$$y=\frac{-10x}{3}$$

Subtract \frac{-10x}{3} from both sides.

$$y-\frac{-10x}{3}=0$$

Multiply both sides of the equation by 3.

$$3y+10x=0$$

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.

$$x+30y=0,10x+3y=0$$

To make x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 1.

$$10x+10\times 30y=0,10x+3y=0$$

Simplify.

$$10x+300y=0,10x+3y=0$$

Subtract 10x+3y=0 from 10x+300y=0 by subtracting like terms on each side of the equal sign.

$$10x-10x+300y-3y=0$$

Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.

$$300y-3y=0$$

Add 300y to -3y.

$$297y=0$$

Divide both sides by 297.

$$y=0$$

Substitute 0 for y in 10x+3y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x=0$$

Divide both sides by 10.

$$x=0$$

The system is now solved.

$$x=0,y=0$$