Solve {l}{x/4-3/8y=3}{5/3x-y/2=12}

$$\left. \begin{array} { l } { \frac { x } { 4 } - \frac { 3 } { 8 } y = 3 } \\ { \frac { 5 } { 3 } x - \frac { y } { 2 } = 12 } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=-4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2x-3y=24$$

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

$$10x-3y=72$$

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.

$$2x-3y=24,10x-3y=72$$

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, 10x-3y=72.

$$10\left(\frac{3}{2}y+12\right)-3y=72$$

Multiply 10 times \frac{3y}{2}+12.

$$15y+120-3y=72$$

Add 15y to -3y.

$$12y+120=72$$

Subtract 120 from both sides of the equation.

$$12y=-48$$

Divide both sides by 12.

$$y=-4$$

Substitute -4 for y in x=\frac{3}{2}y+12. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\left(-4\right)+12$$

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

$$x=-6+12$$

Add 12 to -6.

$$x=6$$

The system is now solved.

$$x=6,y=-4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2x-3y=24$$

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

$$10x-3y=72$$

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

$$2x-3y=24,10x-3y=72$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}24\\72\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\10&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right))\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right))\left(\begin{matrix}24\\72\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right))\left(\begin{matrix}24\\72\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\10&-3\end{matrix}\right))\left(\begin{matrix}24\\72\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{2\left(-3\right)-\left(-3\times 10\right)}&-\frac{-3}{2\left(-3\right)-\left(-3\times 10\right)}\\-\frac{10}{2\left(-3\right)-\left(-3\times 10\right)}&\frac{2}{2\left(-3\right)-\left(-3\times 10\right)}\end{matrix}\right)\left(\begin{matrix}24\\72\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}&\frac{1}{8}\\-\frac{5}{12}&\frac{1}{12}\end{matrix}\right)\left(\begin{matrix}24\\72\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{8}\times 24+\frac{1}{8}\times 72\\-\frac{5}{12}\times 24+\frac{1}{12}\times 72\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\-4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=-4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 4,8.

$$2x-3y=24$$

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

$$10x-3y=72$$

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.

$$2x-3y=24,10x-3y=72$$

Subtract 10x-3y=72 from 2x-3y=24 by subtracting like terms on each side of the equal sign.

$$2x-10x-3y+3y=24-72$$

Add -3y to 3y. Terms -3y and 3y cancel out, leaving an equation with only one variable that can be solved.

$$2x-10x=24-72$$

Add 2x to -10x.

$$-8x=24-72$$

Add 24 to -72.

$$-8x=-48$$

Divide both sides by -8.

$$x=6$$

Substitute 6 for x in 10x-3y=72. Because the resulting equation contains only one variable, you can solve for y directly.

$$10\times 6-3y=72$$

Multiply 10 times 6.

$$60-3y=72$$

Subtract 60 from both sides of the equation.

$$-3y=12$$

Divide both sides by -3.

$$y=-4$$

The system is now solved.

$$x=6,y=-4$$