Solve {l}{4x-y-6/3=18}{1/3y+4x-12/2=8}

$$\left\{ \begin{array} { l } { 4 x - \frac { y - 6 } { 3 } = 18 } \\ { \frac { 1 } { 3 } y + \frac { 4 x - 12 } { 2 } = 8 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=12$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 3.

$$12x-\left(y-6\right)=54$$

To find the opposite of y-6, find the opposite of each term.

$$12x-y+6=54$$

Subtract 6 from both sides.

$$12x-y=54-6$$

Subtract 6 from 54 to get 48.

$$12x-y=48$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2y+3\left(4x-12\right)=48$$

Use the distributive property to multiply 3 by 4x-12.

$$2y+12x-36=48$$

Add 36 to both sides.

$$2y+12x=48+36$$

Add 48 and 36 to get 84.

$$2y+12x=84$$

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.

$$12x-y=48,12x+2y=84$$

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

$$12x-y=48$$

Add y to both sides of the equation.

$$12x=y+48$$

Divide both sides by 12.

$$x=\frac{1}{12}\left(y+48\right)$$

Multiply \frac{1}{12} times y+48.

$$x=\frac{1}{12}y+4$$

Substitute \frac{y}{12}+4 for x in the other equation, 12x+2y=84.

$$12\left(\frac{1}{12}y+4\right)+2y=84$$

Multiply 12 times \frac{y}{12}+4.

$$y+48+2y=84$$

Add y to 2y.

$$3y+48=84$$

Subtract 48 from both sides of the equation.

$$3y=36$$

Divide both sides by 3.

$$y=12$$

Substitute 12 for y in x=\frac{1}{12}y+4. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{1}{12}\times 12+4$$

Multiply \frac{1}{12} times 12.

$$x=1+4$$

Add 4 to 1.

$$x=5$$

The system is now solved.

$$x=5,y=12$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 3.

$$12x-\left(y-6\right)=54$$

To find the opposite of y-6, find the opposite of each term.

$$12x-y+6=54$$

Subtract 6 from both sides.

$$12x-y=54-6$$

Subtract 6 from 54 to get 48.

$$12x-y=48$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2y+3\left(4x-12\right)=48$$

Use the distributive property to multiply 3 by 4x-12.

$$2y+12x-36=48$$

Add 36 to both sides.

$$2y+12x=48+36$$

Add 48 and 36 to get 84.

$$2y+12x=84$$

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

$$12x-y=48,12x+2y=84$$

Write the equations in matrix form.

$$\left(\begin{matrix}12&-1\\12&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}48\\84\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}12&-1\\12&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}12&-1\\12&2\end{matrix}\right))\left(\begin{matrix}12&-1\\12&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-1\\12&2\end{matrix}\right))\left(\begin{matrix}48\\84\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-1\\12&2\end{matrix}\right))\left(\begin{matrix}48\\84\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}12&-1\\12&2\end{matrix}\right))\left(\begin{matrix}48\\84\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{12\times 2-\left(-12\right)}&-\frac{-1}{12\times 2-\left(-12\right)}\\-\frac{12}{12\times 2-\left(-12\right)}&\frac{12}{12\times 2-\left(-12\right)}\end{matrix}\right)\left(\begin{matrix}48\\84\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{18}&\frac{1}{36}\\-\frac{1}{3}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}48\\84\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{18}\times 48+\frac{1}{36}\times 84\\-\frac{1}{3}\times 48+\frac{1}{3}\times 84\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\12\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=12$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 3.

$$12x-\left(y-6\right)=54$$

To find the opposite of y-6, find the opposite of each term.

$$12x-y+6=54$$

Subtract 6 from both sides.

$$12x-y=54-6$$

Subtract 6 from 54 to get 48.

$$12x-y=48$$

Consider the second equation. Multiply both sides of the equation by 6, the least common multiple of 3,2.

$$2y+3\left(4x-12\right)=48$$

Use the distributive property to multiply 3 by 4x-12.

$$2y+12x-36=48$$

Add 36 to both sides.

$$2y+12x=48+36$$

Add 48 and 36 to get 84.

$$2y+12x=84$$

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.

$$12x-y=48,12x+2y=84$$

Subtract 12x+2y=84 from 12x-y=48 by subtracting like terms on each side of the equal sign.

$$12x-12x-y-2y=48-84$$

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

$$-y-2y=48-84$$

Add -y to -2y.

$$-3y=48-84$$

Add 48 to -84.

$$-3y=-36$$

Divide both sides by -3.

$$y=12$$

Substitute 12 for y in 12x+2y=84. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+2\times 12=84$$

Multiply 2 times 12.

$$12x+24=84$$

Subtract 24 from both sides of the equation.

$$12x=60$$

Divide both sides by 12.

$$x=5$$

The system is now solved.

$$x=5,y=12$$