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x+y=240,200x+100y=33000
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=240
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+240
Subtract y from both sides of the equation.
200\left(-y+240\right)+100y=33000
Substitute -y+240 for x in the other equation, 200x+100y=33000.
-200y+48000+100y=33000
Multiply 200 times -y+240.
-100y+48000=33000
Add -200y to 100y.
-100y=-15000
Subtract 48000 from both sides of the equation.
y=150
Divide both sides by -100.
x=-150+240
Substitute 150 for y in x=-y+240. Because the resulting equation contains only one variable, you can solve for x directly.
x=90
Add 240 to -150.
x=90,y=150
The system is now solved.
x+y=240,200x+100y=33000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\200&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}240\\33000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\200&100\end{matrix}\right))\left(\begin{matrix}1&1\\200&100\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\200&100\end{matrix}\right))\left(\begin{matrix}240\\33000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\200&100\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\\200&100\end{matrix}\right))\left(\begin{matrix}240\\33000\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\\200&100\end{matrix}\right))\left(\begin{matrix}240\\33000\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{100}{100-200}&-\frac{1}{100-200}\\-\frac{200}{100-200}&\frac{1}{100-200}\end{matrix}\right)\left(\begin{matrix}240\\33000\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}-1&\frac{1}{100}\\2&-\frac{1}{100}\end{matrix}\right)\left(\begin{matrix}240\\33000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-240+\frac{1}{100}\times 33000\\2\times 240-\frac{1}{100}\times 33000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}90\\150\end{matrix}\right)
Do the arithmetic.
x=90,y=150
Extract the matrix elements x and y.
x+y=240,200x+100y=33000
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.
200x+200y=200\times 240,200x+100y=33000
To make x and 200x equal, multiply all terms on each side of the first equation by 200 and all terms on each side of the second by 1.
200x+200y=48000,200x+100y=33000
Simplify.
200x-200x+200y-100y=48000-33000
Subtract 200x+100y=33000 from 200x+200y=48000 by subtracting like terms on each side of the equal sign.
200y-100y=48000-33000
Add 200x to -200x. Terms 200x and -200x cancel out, leaving an equation with only one variable that can be solved.
100y=48000-33000
Add 200y to -100y.
100y=15000
Add 48000 to -33000.
y=150
Divide both sides by 100.
200x+100\times 150=33000
Substitute 150 for y in 200x+100y=33000. Because the resulting equation contains only one variable, you can solve for x directly.
200x+15000=33000
Multiply 100 times 150.
200x=18000
Subtract 15000 from both sides of the equation.
x=90
Divide both sides by 200.
x=90,y=150
The system is now solved.