Solve {l}{x=3y}{x-6=2y+12}

$$\left\{ \begin{array} { l } { x = 3 y } \\ { x - 6 = 2 y + 12 } \end{array} \right.$$

Solve for x, y

$$x=54$$

$$y=18$$

Steps Using Substitution

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

$$x-3y=0$$

Consider the second equation. Subtract 2y from both sides.

$$x-6-2y=12$$

Add 6 to both sides.

$$x-2y=12+6$$

Add 12 and 6 to get 18.

$$x-2y=18$$

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.

$$x-3y=0,x-2y=18$$

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

$$x-3y=0$$

Add 3y to both sides of the equation.

$$x=3y$$

Substitute 3y for x in the other equation, x-2y=18.

$$3y-2y=18$$

Add 3y to -2y.

$$y=18$$

Substitute 18 for y in x=3y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=3\times 18$$

Multiply 3 times 18.

$$x=54$$

The system is now solved.

$$x=54,y=18$$

Steps Using Matrices

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

$$x-3y=0$$

Consider the second equation. Subtract 2y from both sides.

$$x-6-2y=12$$

Add 6 to both sides.

$$x-2y=12+6$$

Add 12 and 6 to get 18.

$$x-2y=18$$

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

$$x-3y=0,x-2y=18$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\18\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-3\\1&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}0\\18\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}0\\18\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-3\\1&-2\end{matrix}\right))\left(\begin{matrix}0\\18\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-2-\left(-3\right)}&-\frac{-3}{-2-\left(-3\right)}\\-\frac{1}{-2-\left(-3\right)}&\frac{1}{-2-\left(-3\right)}\end{matrix}\right)\left(\begin{matrix}0\\18\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-2&3\\-1&1\end{matrix}\right)\left(\begin{matrix}0\\18\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\times 18\\18\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}54\\18\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=54,y=18$$

Steps Using Elimination

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

$$x-3y=0$$

Consider the second equation. Subtract 2y from both sides.

$$x-6-2y=12$$

Add 6 to both sides.

$$x-2y=12+6$$

Add 12 and 6 to get 18.

$$x-2y=18$$

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.

$$x-3y=0,x-2y=18$$

Subtract x-2y=18 from x-3y=0 by subtracting like terms on each side of the equal sign.

$$x-x-3y+2y=-18$$

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

$$-3y+2y=-18$$

Add -3y to 2y.

$$-y=-18$$

Divide both sides by -1.

$$y=18$$

Substitute 18 for y in x-2y=18. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-2\times 18=18$$

Multiply -2 times 18.

$$x-36=18$$

Add 36 to both sides of the equation.

$$x=54$$

The system is now solved.

$$x=54,y=18$$