Solve {l}{x/8-3y/2=-5}{y/4+9x/2=25}

$$\left.\begin{array} { l } { \frac { x } { 8 } - \frac { 3 y } { 2 } = - 5 } \\ { \frac { y } { 4 } + \frac { 9 x } { 2 } = 25 } \end{array} \right\}$$

Solve for x, y

$$x = \frac{1160}{217} = 5\frac{75}{217} \approx 5.34562212$$

$$y = \frac{820}{217} = 3\frac{169}{217} \approx 3.778801843$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 8,2.

$$x-4\times 3y=-40$$

Multiply -4 and 3 to get -12.

$$x-12y=-40$$

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$y+2\times 9x=100$$

Multiply 2 and 9 to get 18.

$$y+18x=100$$

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.

$$x-12y=-40,18x+y=100$$

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

$$x-12y=-40$$

Add 12y to both sides of the equation.

$$x=12y-40$$

Substitute 12y-40 for x in the other equation, 18x+y=100.

$$18\left(12y-40\right)+y=100$$

Multiply 18 times 12y-40.

$$216y-720+y=100$$

Add 216y to y.

$$217y-720=100$$

Add 720 to both sides of the equation.

$$217y=820$$

Divide both sides by 217.

$$y=\frac{820}{217}$$

Substitute \frac{820}{217} for y in x=12y-40. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=12\times \frac{820}{217}-40$$

Multiply 12 times \frac{820}{217}.

$$x=\frac{9840}{217}-40$$

Add -40 to \frac{9840}{217}.

$$x=\frac{1160}{217}$$

The system is now solved.

$$x=\frac{1160}{217},y=\frac{820}{217}$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 8,2.

$$x-4\times 3y=-40$$

Multiply -4 and 3 to get -12.

$$x-12y=-40$$

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$y+2\times 9x=100$$

Multiply 2 and 9 to get 18.

$$y+18x=100$$

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

$$x-12y=-40,18x+y=100$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-12\\18&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-40\\100\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-12\\18&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-12\\18&1\end{matrix}\right))\left(\begin{matrix}1&-12\\18&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-12\\18&1\end{matrix}\right))\left(\begin{matrix}-40\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-12\\18&1\end{matrix}\right))\left(\begin{matrix}-40\\100\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-12\\18&1\end{matrix}\right))\left(\begin{matrix}-40\\100\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{1-\left(-12\times 18\right)}&-\frac{-12}{1-\left(-12\times 18\right)}\\-\frac{18}{1-\left(-12\times 18\right)}&\frac{1}{1-\left(-12\times 18\right)}\end{matrix}\right)\left(\begin{matrix}-40\\100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{217}&\frac{12}{217}\\-\frac{18}{217}&\frac{1}{217}\end{matrix}\right)\left(\begin{matrix}-40\\100\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{217}\left(-40\right)+\frac{12}{217}\times 100\\-\frac{18}{217}\left(-40\right)+\frac{1}{217}\times 100\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1160}{217}\\\frac{820}{217}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1160}{217},y=\frac{820}{217}$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 8, the least common multiple of 8,2.

$$x-4\times 3y=-40$$

Multiply -4 and 3 to get -12.

$$x-12y=-40$$

Consider the second equation. Multiply both sides of the equation by 4, the least common multiple of 4,2.

$$y+2\times 9x=100$$

Multiply 2 and 9 to get 18.

$$y+18x=100$$

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.

$$x-12y=-40,18x+y=100$$

To make x and 18x equal, multiply all terms on each side of the first equation by 18 and all terms on each side of the second by 1.

$$18x+18\left(-12\right)y=18\left(-40\right),18x+y=100$$

Simplify.

$$18x-216y=-720,18x+y=100$$

Subtract 18x+y=100 from 18x-216y=-720 by subtracting like terms on each side of the equal sign.

$$18x-18x-216y-y=-720-100$$

Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.

$$-216y-y=-720-100$$

Add -216y to -y.

$$-217y=-720-100$$

Add -720 to -100.

$$-217y=-820$$

Divide both sides by -217.

$$y=\frac{820}{217}$$

Substitute \frac{820}{217} for y in 18x+y=100. Because the resulting equation contains only one variable, you can solve for x directly.

$$18x+\frac{820}{217}=100$$

Subtract \frac{820}{217} from both sides of the equation.

$$18x=\frac{20880}{217}$$

Divide both sides by 18.

$$x=\frac{1160}{217}$$

The system is now solved.

$$x=\frac{1160}{217},y=\frac{820}{217}$$