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x+2y=2500,2x+y=2450
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+2y=2500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-2y+2500
Subtract 2y from both sides of the equation.
2\left(-2y+2500\right)+y=2450
Substitute -2y+2500 for x in the other equation, 2x+y=2450.
-4y+5000+y=2450
Multiply 2 times -2y+2500.
-3y+5000=2450
Add -4y to y.
-3y=-2550
Subtract 5000 from both sides of the equation.
y=850
Divide both sides by -3.
x=-2\times 850+2500
Substitute 850 for y in x=-2y+2500. Because the resulting equation contains only one variable, you can solve for x directly.
x=-1700+2500
Multiply -2 times 850.
x=800
Add 2500 to -1700.
x=800,y=850
The system is now solved.
x+2y=2500,2x+y=2450
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2500\\2450\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}1&2\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&2\\2&1\end{matrix}\right))\left(\begin{matrix}2500\\2450\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&2\\2&1\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&2\\2&1\end{matrix}\right))\left(\begin{matrix}2500\\2450\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&2\\2&1\end{matrix}\right))\left(\begin{matrix}2500\\2450\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{1}{1-2\times 2}&-\frac{2}{1-2\times 2}\\-\frac{2}{1-2\times 2}&\frac{1}{1-2\times 2}\end{matrix}\right)\left(\begin{matrix}2500\\2450\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{1}{3}&\frac{2}{3}\\\frac{2}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}2500\\2450\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 2500+\frac{2}{3}\times 2450\\\frac{2}{3}\times 2500-\frac{1}{3}\times 2450\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}800\\850\end{matrix}\right)
Do the arithmetic.
x=800,y=850
Extract the matrix elements x and y.
x+2y=2500,2x+y=2450
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.
2x+2\times 2y=2\times 2500,2x+y=2450
To make x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.
2x+4y=5000,2x+y=2450
Simplify.
2x-2x+4y-y=5000-2450
Subtract 2x+y=2450 from 2x+4y=5000 by subtracting like terms on each side of the equal sign.
4y-y=5000-2450
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
3y=5000-2450
Add 4y to -y.
3y=2550
Add 5000 to -2450.
y=850
Divide both sides by 3.
2x+850=2450
Substitute 850 for y in 2x+y=2450. Because the resulting equation contains only one variable, you can solve for x directly.
2x=1600
Subtract 850 from both sides of the equation.
x=800
Divide both sides by 2.
x=800,y=850
The system is now solved.