Solve {l}{x+y=130}{20x+5y=1925}

$$\left. \begin{array} { l } { x + y = 130 } \\ { 20 x + 5 y = 1925 } \end{array} \right.$$

Solve for x, y

$$x=85$$

$$y=45$$

Steps Using Substitution

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.

$$x+y=130,20x+5y=1925$$

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, 20x+5y=1925.

$$20\left(-y+130\right)+5y=1925$$

Multiply 20 times -y+130.

$$-20y+2600+5y=1925$$

Add -20y to 5y.

$$-15y+2600=1925$$

Subtract 2600 from both sides of the equation.

$$-15y=-675$$

Divide both sides by -15.

$$y=45$$

Substitute 45 for y in x=-y+130. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-45+130$$

Add 130 to -45.

$$x=85$$

The system is now solved.

$$x=85,y=45$$

Steps Using Matrices

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

$$x+y=130,20x+5y=1925$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\20&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130\\1925\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\20&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\20&5\end{matrix}\right))\left(\begin{matrix}1&1\\20&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\20&5\end{matrix}\right))\left(\begin{matrix}130\\1925\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\20&5\end{matrix}\right))\left(\begin{matrix}130\\1925\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\20&5\end{matrix}\right))\left(\begin{matrix}130\\1925\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5-20}&-\frac{1}{5-20}\\-\frac{20}{5-20}&\frac{1}{5-20}\end{matrix}\right)\left(\begin{matrix}130\\1925\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&\frac{1}{15}\\\frac{4}{3}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}130\\1925\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 130+\frac{1}{15}\times 1925\\\frac{4}{3}\times 130-\frac{1}{15}\times 1925\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}85\\45\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=85,y=45$$

Steps Using Elimination

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.

$$x+y=130,20x+5y=1925$$

To make x and 20x equal, multiply all terms on each side of the first equation by 20 and all terms on each side of the second by 1.

$$20x+20y=20\times 130,20x+5y=1925$$

Simplify.

$$20x+20y=2600,20x+5y=1925$$

Subtract 20x+5y=1925 from 20x+20y=2600 by subtracting like terms on each side of the equal sign.

$$20x-20x+20y-5y=2600-1925$$

Add 20x to -20x. Terms 20x and -20x cancel out, leaving an equation with only one variable that can be solved.

$$20y-5y=2600-1925$$

Add 20y to -5y.

$$15y=2600-1925$$

Add 2600 to -1925.

$$15y=675$$

Divide both sides by 15.

$$y=45$$

Substitute 45 for y in 20x+5y=1925. Because the resulting equation contains only one variable, you can solve for x directly.

$$20x+5\times 45=1925$$

Multiply 5 times 45.

$$20x+225=1925$$

Subtract 225 from both sides of the equation.

$$20x=1700$$

Divide both sides by 20.

$$x=85$$

The system is now solved.

$$x=85,y=45$$