Solve {c}{x+y=35}{40x+35y=1300}

$$\left. \begin{array} { c } { x + y = 35 } \\ { 40 x + 35 y = 1300 } \end{array} \right.$$

Solve for x, y

$$x=15$$

$$y=20$$

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.

$$x+y=35,40x+35y=1300$$

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

$$x+y=35$$

Subtract y from both sides of the equation.

$$x=-y+35$$

Substitute -y+35 for x in the other equation, 40x+35y=1300.

$$40\left(-y+35\right)+35y=1300$$

Multiply 40 times -y+35.

$$-40y+1400+35y=1300$$

Add -40y to 35y.

$$-5y+1400=1300$$

Subtract 1400 from both sides of the equation.

$$-5y=-100$$

Divide both sides by -5.

$$y=20$$

Substitute 20 for y in x=-y+35. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-20+35$$

Add 35 to -20.

$$x=15$$

The system is now solved.

$$x=15,y=20$$

Steps Using Matrices

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

$$x+y=35,40x+35y=1300$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\40&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\1300\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\40&35\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\40&35\end{matrix}\right))\left(\begin{matrix}1&1\\40&35\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\40&35\end{matrix}\right))\left(\begin{matrix}35\\1300\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\40&35\end{matrix}\right))\left(\begin{matrix}35\\1300\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\40&35\end{matrix}\right))\left(\begin{matrix}35\\1300\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{35}{35-40}&-\frac{1}{35-40}\\-\frac{40}{35-40}&\frac{1}{35-40}\end{matrix}\right)\left(\begin{matrix}35\\1300\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-7&\frac{1}{5}\\8&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}35\\1300\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-7\times 35+\frac{1}{5}\times 1300\\8\times 35-\frac{1}{5}\times 1300\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=15,y=20$$

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.

$$x+y=35,40x+35y=1300$$

To make x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 1.

$$40x+40y=40\times 35,40x+35y=1300$$

Simplify.

$$40x+40y=1400,40x+35y=1300$$

Subtract 40x+35y=1300 from 40x+40y=1400 by subtracting like terms on each side of the equal sign.

$$40x-40x+40y-35y=1400-1300$$

Add 40x to -40x. Terms 40x and -40x cancel out, leaving an equation with only one variable that can be solved.

$$40y-35y=1400-1300$$

Add 40y to -35y.

$$5y=1400-1300$$

Add 1400 to -1300.

$$5y=100$$

Divide both sides by 5.

$$y=20$$

Substitute 20 for y in 40x+35y=1300. Because the resulting equation contains only one variable, you can solve for x directly.

$$40x+35\times 20=1300$$

Multiply 35 times 20.

$$40x+700=1300$$

Subtract 700 from both sides of the equation.

$$40x=600$$

Divide both sides by 40.

$$x=15$$

The system is now solved.

$$x=15,y=20$$