Solve {l}{x+y=85}{65x+50y=5075}

$$\left. \begin{array} { l } { x + y = 85 } \\ { 65 x + 50 y = 5075 } \end{array} \right.$$

Solve for x, y

$$x=55$$

$$y=30$$

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.

$$x+y=85,65x+50y=5075$$

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

$$x+y=85$$

Subtract y from both sides of the equation.

$$x=-y+85$$

Substitute -y+85 for x in the other equation, 65x+50y=5075.

$$65\left(-y+85\right)+50y=5075$$

Multiply 65 times -y+85.

$$-65y+5525+50y=5075$$

Add -65y to 50y.

$$-15y+5525=5075$$

Subtract 5525 from both sides of the equation.

$$-15y=-450$$

Divide both sides by -15.

$$y=30$$

Substitute 30 for y in x=-y+85. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-30+85$$

Add 85 to -30.

$$x=55$$

The system is now solved.

$$x=55,y=30$$

Steps Using Matrices

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

$$x+y=85,65x+50y=5075$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\65&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}85\\5075\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\65&50\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\65&50\end{matrix}\right))\left(\begin{matrix}1&1\\65&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\65&50\end{matrix}\right))\left(\begin{matrix}85\\5075\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\65&50\end{matrix}\right))\left(\begin{matrix}85\\5075\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\65&50\end{matrix}\right))\left(\begin{matrix}85\\5075\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{50}{50-65}&-\frac{1}{50-65}\\-\frac{65}{50-65}&\frac{1}{50-65}\end{matrix}\right)\left(\begin{matrix}85\\5075\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{10}{3}&\frac{1}{15}\\\frac{13}{3}&-\frac{1}{15}\end{matrix}\right)\left(\begin{matrix}85\\5075\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{10}{3}\times 85+\frac{1}{15}\times 5075\\\frac{13}{3}\times 85-\frac{1}{15}\times 5075\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}55\\30\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=55,y=30$$

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.

$$x+y=85,65x+50y=5075$$

To make x and 65x equal, multiply all terms on each side of the first equation by 65 and all terms on each side of the second by 1.

$$65x+65y=65\times 85,65x+50y=5075$$

Simplify.

$$65x+65y=5525,65x+50y=5075$$

Subtract 65x+50y=5075 from 65x+65y=5525 by subtracting like terms on each side of the equal sign.

$$65x-65x+65y-50y=5525-5075$$

Add 65x to -65x. Terms 65x and -65x cancel out, leaving an equation with only one variable that can be solved.

$$65y-50y=5525-5075$$

Add 65y to -50y.

$$15y=5525-5075$$

Add 5525 to -5075.

$$15y=450$$

Divide both sides by 15.

$$y=30$$

Substitute 30 for y in 65x+50y=5075. Because the resulting equation contains only one variable, you can solve for x directly.

$$65x+50\times 30=5075$$

Multiply 50 times 30.

$$65x+1500=5075$$

Subtract 1500 from both sides of the equation.

$$65x=3575$$

Divide both sides by 65.

$$x=55$$

The system is now solved.

$$x=55,y=30$$