Skip to main content
Solve for x, y
Tick mark Image
Graph

Similar Problems from Web Search

Share

x+y=8,50x+80y=550
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=8
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+8
Subtract y from both sides of the equation.
50\left(-y+8\right)+80y=550
Substitute -y+8 for x in the other equation, 50x+80y=550.
-50y+400+80y=550
Multiply 50 times -y+8.
30y+400=550
Add -50y to 80y.
30y=150
Subtract 400 from both sides of the equation.
y=5
Divide both sides by 30.
x=-5+8
Substitute 5 for y in x=-y+8. Because the resulting equation contains only one variable, you can solve for x directly.
x=3
Add 8 to -5.
x=3,y=5
The system is now solved.
x+y=8,50x+80y=550
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\50&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\550\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\50&80\end{matrix}\right))\left(\begin{matrix}1&1\\50&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\50&80\end{matrix}\right))\left(\begin{matrix}8\\550\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\50&80\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\50&80\end{matrix}\right))\left(\begin{matrix}8\\550\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\50&80\end{matrix}\right))\left(\begin{matrix}8\\550\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{80}{80-50}&-\frac{1}{80-50}\\-\frac{50}{80-50}&\frac{1}{80-50}\end{matrix}\right)\left(\begin{matrix}8\\550\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{8}{3}&-\frac{1}{30}\\-\frac{5}{3}&\frac{1}{30}\end{matrix}\right)\left(\begin{matrix}8\\550\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{3}\times 8-\frac{1}{30}\times 550\\-\frac{5}{3}\times 8+\frac{1}{30}\times 550\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}3\\5\end{matrix}\right)
Do the arithmetic.
x=3,y=5
Extract the matrix elements x and y.
x+y=8,50x+80y=550
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.
50x+50y=50\times 8,50x+80y=550
To make x and 50x equal, multiply all terms on each side of the first equation by 50 and all terms on each side of the second by 1.
50x+50y=400,50x+80y=550
Simplify.
50x-50x+50y-80y=400-550
Subtract 50x+80y=550 from 50x+50y=400 by subtracting like terms on each side of the equal sign.
50y-80y=400-550
Add 50x to -50x. Terms 50x and -50x cancel out, leaving an equation with only one variable that can be solved.
-30y=400-550
Add 50y to -80y.
-30y=-150
Add 400 to -550.
y=5
Divide both sides by -30.
50x+80\times 5=550
Substitute 5 for y in 50x+80y=550. Because the resulting equation contains only one variable, you can solve for x directly.
50x+400=550
Multiply 80 times 5.
50x=150
Subtract 400 from both sides of the equation.
x=3
Divide both sides by 50.
x=3,y=5
The system is now solved.