Solve {l}{x+y=80}{35x+85y=80(x+y)}

$$\left. \begin{array} { l } { x + y = 80 } \\ { 35 x + 85 y = 80 ( x + y ) } \end{array} \right.$$

Solve for x, y

$$x=8$$

$$y=72$$

Steps Using Substitution

Consider the second equation. Use the distributive property to multiply 80 by x+y.

$$35x+85y=80x+80y$$

Subtract 80x from both sides.

$$35x+85y-80x=80y$$

Combine 35x and -80x to get -45x.

$$-45x+85y=80y$$

Subtract 80y from both sides.

$$-45x+85y-80y=0$$

Combine 85y and -80y to get 5y.

$$-45x+5y=0$$

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.

$$x+y=80,-45x+5y=0$$

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

$$x+y=80$$

Subtract y from both sides of the equation.

$$x=-y+80$$

Substitute -y+80 for x in the other equation, -45x+5y=0.

$$-45\left(-y+80\right)+5y=0$$

Multiply -45 times -y+80.

$$45y-3600+5y=0$$

Add 45y to 5y.

$$50y-3600=0$$

Add 3600 to both sides of the equation.

$$50y=3600$$

Divide both sides by 50.

$$y=72$$

Substitute 72 for y in x=-y+80. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-72+80$$

Add 80 to -72.

$$x=8$$

The system is now solved.

$$x=8,y=72$$

Steps Using Matrices

Consider the second equation. Use the distributive property to multiply 80 by x+y.

$$35x+85y=80x+80y$$

Subtract 80x from both sides.

$$35x+85y-80x=80y$$

Combine 35x and -80x to get -45x.

$$-45x+85y=80y$$

Subtract 80y from both sides.

$$-45x+85y-80y=0$$

Combine 85y and -80y to get 5y.

$$-45x+5y=0$$

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

$$x+y=80,-45x+5y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\-45&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-45&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\-45&5\end{matrix}\right))\left(\begin{matrix}1&1\\-45&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-45&5\end{matrix}\right))\left(\begin{matrix}80\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-45&5\end{matrix}\right))\left(\begin{matrix}80\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-45&5\end{matrix}\right))\left(\begin{matrix}80\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{5-\left(-45\right)}&-\frac{1}{5-\left(-45\right)}\\-\frac{-45}{5-\left(-45\right)}&\frac{1}{5-\left(-45\right)}\end{matrix}\right)\left(\begin{matrix}80\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}&-\frac{1}{50}\\\frac{9}{10}&\frac{1}{50}\end{matrix}\right)\left(\begin{matrix}80\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10}\times 80\\\frac{9}{10}\times 80\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\72\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=8,y=72$$

Steps Using Elimination

Consider the second equation. Use the distributive property to multiply 80 by x+y.

$$35x+85y=80x+80y$$

Subtract 80x from both sides.

$$35x+85y-80x=80y$$

Combine 35x and -80x to get -45x.

$$-45x+85y=80y$$

Subtract 80y from both sides.

$$-45x+85y-80y=0$$

Combine 85y and -80y to get 5y.

$$-45x+5y=0$$

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.

$$x+y=80,-45x+5y=0$$

To make x and -45x equal, multiply all terms on each side of the first equation by -45 and all terms on each side of the second by 1.

$$-45x-45y=-45\times 80,-45x+5y=0$$

Simplify.

$$-45x-45y=-3600,-45x+5y=0$$

Subtract -45x+5y=0 from -45x-45y=-3600 by subtracting like terms on each side of the equal sign.

$$-45x+45x-45y-5y=-3600$$

Add -45x to 45x. Terms -45x and 45x cancel out, leaving an equation with only one variable that can be solved.

$$-45y-5y=-3600$$

Add -45y to -5y.

$$-50y=-3600$$

Divide both sides by -50.

$$y=72$$

Substitute 72 for y in -45x+5y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$-45x+5\times 72=0$$

Multiply 5 times 72.

$$-45x+360=0$$

Subtract 360 from both sides of the equation.

$$-45x=-360$$

Divide both sides by -45.

$$x=8$$

The system is now solved.

$$x=8,y=72$$