Solve {l}{x+y=130}{8x=22y*5}

$$\left. \begin{array} { l } { x + y = 130 } \\ { 8 x = 22 y \times 5 } \end{array} \right.$$

Solve for x, y

$$x = \frac{7150}{59} = 121\frac{11}{59} \approx 121.186440678$$

$$y = \frac{520}{59} = 8\frac{48}{59} \approx 8.813559322$$

Steps Using Substitution

Consider the second equation. Multiply 22 and 5 to get 110.

$$8x=110y$$

Subtract 110y from both sides.

$$8x-110y=0$$

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.

$$x+y=130,8x-110y=0$$

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

$$x+y=130$$

Subtract y from both sides of the equation.

$$x=-y+130$$

Substitute -y+130 for x in the other equation, 8x-110y=0.

$$8\left(-y+130\right)-110y=0$$

Multiply 8 times -y+130.

$$-8y+1040-110y=0$$

Add -8y to -110y.

$$-118y+1040=0$$

Subtract 1040 from both sides of the equation.

$$-118y=-1040$$

Divide both sides by -118.

$$y=\frac{520}{59}$$

Substitute \frac{520}{59} for y in x=-y+130. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{520}{59}+130$$

Add 130 to -\frac{520}{59}.

$$x=\frac{7150}{59}$$

The system is now solved.

$$x=\frac{7150}{59},y=\frac{520}{59}$$

Steps Using Matrices

Consider the second equation. Multiply 22 and 5 to get 110.

$$8x=110y$$

Subtract 110y from both sides.

$$8x-110y=0$$

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

$$x+y=130,8x-110y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\8&-110\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\8&-110\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\8&-110\end{matrix}\right))\left(\begin{matrix}1&1\\8&-110\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\8&-110\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\8&-110\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\8&-110\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{110}{-110-8}&-\frac{1}{-110-8}\\-\frac{8}{-110-8}&\frac{1}{-110-8}\end{matrix}\right)\left(\begin{matrix}130\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{55}{59}&\frac{1}{118}\\\frac{4}{59}&-\frac{1}{118}\end{matrix}\right)\left(\begin{matrix}130\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{55}{59}\times 130\\\frac{4}{59}\times 130\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7150}{59}\\\frac{520}{59}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{7150}{59},y=\frac{520}{59}$$

Steps Using Elimination

Consider the second equation. Multiply 22 and 5 to get 110.

$$8x=110y$$

Subtract 110y from both sides.

$$8x-110y=0$$

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.

$$x+y=130,8x-110y=0$$

To make x and 8x equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 1.

$$8x+8y=8\times 130,8x-110y=0$$

Simplify.

$$8x+8y=1040,8x-110y=0$$

Subtract 8x-110y=0 from 8x+8y=1040 by subtracting like terms on each side of the equal sign.

$$8x-8x+8y+110y=1040$$

Add 8x to -8x. Terms 8x and -8x cancel out, leaving an equation with only one variable that can be solved.

$$8y+110y=1040$$

Add 8y to 110y.

$$118y=1040$$

Divide both sides by 118.

$$y=\frac{520}{59}$$

Substitute \frac{520}{59} for y in 8x-110y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$8x-110\times \frac{520}{59}=0$$

Multiply -110 times \frac{520}{59}.

$$8x-\frac{57200}{59}=0$$

Add \frac{57200}{59} to both sides of the equation.

$$8x=\frac{57200}{59}$$

Divide both sides by 8.

$$x=\frac{7150}{59}$$

The system is now solved.

$$x=\frac{7150}{59},y=\frac{520}{59}$$