Solve {l}{3(y-6)=x-6}{x=2y}

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

Solve for y, x

$$x=24$$

$$y=12$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 3 by y-6.

$$3y-18=x-6$$

Subtract x from both sides.

$$3y-18-x=-6$$

Add 18 to both sides.

$$3y-x=-6+18$$

Add -6 and 18 to get 12.

$$3y-x=12$$

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

$$x-2y=0$$

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.

$$3y-x=12,-2y+x=0$$

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

$$3y-x=12$$

Add x to both sides of the equation.

$$3y=x+12$$

Divide both sides by 3.

$$y=\frac{1}{3}\left(x+12\right)$$

Multiply \frac{1}{3} times x+12.

$$y=\frac{1}{3}x+4$$

Substitute \frac{x}{3}+4 for y in the other equation, -2y+x=0.

$$-2\left(\frac{1}{3}x+4\right)+x=0$$

Multiply -2 times \frac{x}{3}+4.

$$-\frac{2}{3}x-8+x=0$$

Add -\frac{2x}{3} to x.

$$\frac{1}{3}x-8=0$$

Add 8 to both sides of the equation.

$$\frac{1}{3}x=8$$

Multiply both sides by 3.

$$x=24$$

Substitute 24 for x in y=\frac{1}{3}x+4. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{1}{3}\times 24+4$$

Multiply \frac{1}{3} times 24.

$$y=8+4$$

Add 4 to 8.

$$y=12$$

The system is now solved.

$$y=12,x=24$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 3 by y-6.

$$3y-18=x-6$$

Subtract x from both sides.

$$3y-18-x=-6$$

Add 18 to both sides.

$$3y-x=-6+18$$

Add -6 and 18 to get 12.

$$3y-x=12$$

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

$$x-2y=0$$

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

$$3y-x=12,-2y+x=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}12\\0\end{matrix}\right)$$

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

$$inverse(\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right))\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}3&-1\\-2&1\end{matrix}\right))\left(\begin{matrix}12\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{3-\left(-\left(-2\right)\right)}&-\frac{-1}{3-\left(-\left(-2\right)\right)}\\-\frac{-2}{3-\left(-\left(-2\right)\right)}&\frac{3}{3-\left(-\left(-2\right)\right)}\end{matrix}\right)\left(\begin{matrix}12\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}1&1\\2&3\end{matrix}\right)\left(\begin{matrix}12\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}12\\2\times 12\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}12\\24\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=12,x=24$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 3 by y-6.

$$3y-18=x-6$$

Subtract x from both sides.

$$3y-18-x=-6$$

Add 18 to both sides.

$$3y-x=-6+18$$

Add -6 and 18 to get 12.

$$3y-x=12$$

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

$$x-2y=0$$

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.

$$3y-x=12,-2y+x=0$$

To make 3y and -2y equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 3.

$$-2\times 3y-2\left(-1\right)x=-2\times 12,3\left(-2\right)y+3x=0$$

Simplify.

$$-6y+2x=-24,-6y+3x=0$$

Subtract -6y+3x=0 from -6y+2x=-24 by subtracting like terms on each side of the equal sign.

$$-6y+6y+2x-3x=-24$$

Add -6y to 6y. Terms -6y and 6y cancel out, leaving an equation with only one variable that can be solved.

$$2x-3x=-24$$

Add 2x to -3x.

$$-x=-24$$

Divide both sides by -1.

$$x=24$$

Substitute 24 for x in -2y+x=0. Because the resulting equation contains only one variable, you can solve for y directly.

$$-2y+24=0$$

Subtract 24 from both sides of the equation.

$$-2y=-24$$

Divide both sides by -2.

$$y=12$$

The system is now solved.

$$y=12,x=24$$