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x+y=88,60x+80y=6500
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=88
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+88
Subtract y from both sides of the equation.
60\left(-y+88\right)+80y=6500
Substitute -y+88 for x in the other equation, 60x+80y=6500.
-60y+5280+80y=6500
Multiply 60 times -y+88.
20y+5280=6500
Add -60y to 80y.
20y=1220
Subtract 5280 from both sides of the equation.
y=61
Divide both sides by 20.
x=-61+88
Substitute 61 for y in x=-y+88. Because the resulting equation contains only one variable, you can solve for x directly.
x=27
Add 88 to -61.
x=27,y=61
The system is now solved.
x+y=88,60x+80y=6500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\60&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}88\\6500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\60&80\end{matrix}\right))\left(\begin{matrix}1&1\\60&80\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\60&80\end{matrix}\right))\left(\begin{matrix}88\\6500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\60&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\\60&80\end{matrix}\right))\left(\begin{matrix}88\\6500\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\\60&80\end{matrix}\right))\left(\begin{matrix}88\\6500\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}{80-60}&-\frac{1}{80-60}\\-\frac{60}{80-60}&\frac{1}{80-60}\end{matrix}\right)\left(\begin{matrix}88\\6500\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}4&-\frac{1}{20}\\-3&\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}88\\6500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}4\times 88-\frac{1}{20}\times 6500\\-3\times 88+\frac{1}{20}\times 6500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}27\\61\end{matrix}\right)
Do the arithmetic.
x=27,y=61
Extract the matrix elements x and y.
x+y=88,60x+80y=6500
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.
60x+60y=60\times 88,60x+80y=6500
To make x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 1.
60x+60y=5280,60x+80y=6500
Simplify.
60x-60x+60y-80y=5280-6500
Subtract 60x+80y=6500 from 60x+60y=5280 by subtracting like terms on each side of the equal sign.
60y-80y=5280-6500
Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.
-20y=5280-6500
Add 60y to -80y.
-20y=-1220
Add 5280 to -6500.
y=61
Divide both sides by -20.
60x+80\times 61=6500
Substitute 61 for y in 60x+80y=6500. Because the resulting equation contains only one variable, you can solve for x directly.
60x+4880=6500
Multiply 80 times 61.
60x=1620
Subtract 4880 from both sides of the equation.
x=27
Divide both sides by 60.
x=27,y=61
The system is now solved.