Solve {l}{x/3-y/2=4}{x+y/12=2}

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

Solve for x, y

$$x = \frac{48}{19} = 2\frac{10}{19} \approx 2.526315789$$

$$y = -\frac{120}{19} = -6\frac{6}{19} \approx -6.315789474$$

Steps Using Substitution

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

$$2x-3y=24$$

Consider the second equation. Multiply both sides of the equation by 12.

$$12x+y=24$$

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.

$$2x-3y=24,12x+y=24$$

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

$$2x-3y=24$$

Add 3y to both sides of the equation.

$$2x=3y+24$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(3y+24\right)$$

Multiply \frac{1}{2} times 24+3y.

$$x=\frac{3}{2}y+12$$

Substitute \frac{3y}{2}+12 for x in the other equation, 12x+y=24.

$$12\left(\frac{3}{2}y+12\right)+y=24$$

Multiply 12 times \frac{3y}{2}+12.

$$18y+144+y=24$$

Add 18y to y.

$$19y+144=24$$

Subtract 144 from both sides of the equation.

$$19y=-120$$

Divide both sides by 19.

$$y=-\frac{120}{19}$$

Substitute -\frac{120}{19} for y in x=\frac{3}{2}y+12. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\left(-\frac{120}{19}\right)+12$$

Multiply \frac{3}{2} times -\frac{120}{19} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{180}{19}+12$$

Add 12 to -\frac{180}{19}.

$$x=\frac{48}{19}$$

The system is now solved.

$$x=\frac{48}{19},y=-\frac{120}{19}$$

Steps Using Matrices

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

$$2x-3y=24$$

Consider the second equation. Multiply both sides of the equation by 12.

$$12x+y=24$$

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

$$2x-3y=24,12x+y=24$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\12&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}24\\24\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\12&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\12&1\end{matrix}\right))\left(\begin{matrix}2&-3\\12&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\12&1\end{matrix}\right))\left(\begin{matrix}24\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\12&1\end{matrix}\right))\left(\begin{matrix}24\\24\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\12&1\end{matrix}\right))\left(\begin{matrix}24\\24\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2-\left(-3\times 12\right)}&-\frac{-3}{2-\left(-3\times 12\right)}\\-\frac{12}{2-\left(-3\times 12\right)}&\frac{2}{2-\left(-3\times 12\right)}\end{matrix}\right)\left(\begin{matrix}24\\24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{38}&\frac{3}{38}\\-\frac{6}{19}&\frac{1}{19}\end{matrix}\right)\left(\begin{matrix}24\\24\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{38}\times 24+\frac{3}{38}\times 24\\-\frac{6}{19}\times 24+\frac{1}{19}\times 24\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{48}{19}\\-\frac{120}{19}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{48}{19},y=-\frac{120}{19}$$

Steps Using Elimination

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

$$2x-3y=24$$

Consider the second equation. Multiply both sides of the equation by 12.

$$12x+y=24$$

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.

$$2x-3y=24,12x+y=24$$

To make 2x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 2.

$$12\times 2x+12\left(-3\right)y=12\times 24,2\times 12x+2y=2\times 24$$

Simplify.

$$24x-36y=288,24x+2y=48$$

Subtract 24x+2y=48 from 24x-36y=288 by subtracting like terms on each side of the equal sign.

$$24x-24x-36y-2y=288-48$$

Add 24x to -24x. Terms 24x and -24x cancel out, leaving an equation with only one variable that can be solved.

$$-36y-2y=288-48$$

Add -36y to -2y.

$$-38y=288-48$$

Add 288 to -48.

$$-38y=240$$

Divide both sides by -38.

$$y=-\frac{120}{19}$$

Substitute -\frac{120}{19} for y in 12x+y=24. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x-\frac{120}{19}=24$$

Add \frac{120}{19} to both sides of the equation.

$$12x=\frac{576}{19}$$

Divide both sides by 12.

$$x=\frac{48}{19}$$

The system is now solved.

$$x=\frac{48}{19},y=-\frac{120}{19}$$