Solve {l}{x+y=500}{25x+35y=14500}

$$\left. \begin{array} { l } { x + y = 500 } \\ { 25 x + 35 y = 14500 } \end{array} \right.$$

Solve for x, y

$$x=300$$

$$y=200$$

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.

$$x+y=500,25x+35y=14500$$

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

$$x+y=500$$

Subtract y from both sides of the equation.

$$x=-y+500$$

Substitute -y+500 for x in the other equation, 25x+35y=14500.

$$25\left(-y+500\right)+35y=14500$$

Multiply 25 times -y+500.

$$-25y+12500+35y=14500$$

Add -25y to 35y.

$$10y+12500=14500$$

Subtract 12500 from both sides of the equation.

$$10y=2000$$

Divide both sides by 10.

$$y=200$$

Substitute 200 for y in x=-y+500. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-200+500$$

Add 500 to -200.

$$x=300$$

The system is now solved.

$$x=300,y=200$$

Steps Using Matrices

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

$$x+y=500,25x+35y=14500$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\25&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\14500\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\25&35\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\25&35\end{matrix}\right))\left(\begin{matrix}1&1\\25&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\25&35\end{matrix}\right))\left(\begin{matrix}500\\14500\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\25&35\end{matrix}\right))\left(\begin{matrix}500\\14500\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\25&35\end{matrix}\right))\left(\begin{matrix}500\\14500\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{35-25}&-\frac{1}{35-25}\\-\frac{25}{35-25}&\frac{1}{35-25}\end{matrix}\right)\left(\begin{matrix}500\\14500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}&-\frac{1}{10}\\-\frac{5}{2}&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}500\\14500\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{2}\times 500-\frac{1}{10}\times 14500\\-\frac{5}{2}\times 500+\frac{1}{10}\times 14500\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\200\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=300,y=200$$

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.

$$x+y=500,25x+35y=14500$$

To make x and 25x equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 1.

$$25x+25y=25\times 500,25x+35y=14500$$

Simplify.

$$25x+25y=12500,25x+35y=14500$$

Subtract 25x+35y=14500 from 25x+25y=12500 by subtracting like terms on each side of the equal sign.

$$25x-25x+25y-35y=12500-14500$$

Add 25x to -25x. Terms 25x and -25x cancel out, leaving an equation with only one variable that can be solved.

$$25y-35y=12500-14500$$

Add 25y to -35y.

$$-10y=12500-14500$$

Add 12500 to -14500.

$$-10y=-2000$$

Divide both sides by -10.

$$y=200$$

Substitute 200 for y in 25x+35y=14500. Because the resulting equation contains only one variable, you can solve for x directly.

$$25x+35\times 200=14500$$

Multiply 35 times 200.

$$25x+7000=14500$$

Subtract 7000 from both sides of the equation.

$$25x=7500$$

Divide both sides by 25.

$$x=300$$

The system is now solved.

$$x=300,y=200$$