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x+y=2500,25x+20y=4800
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=2500
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+2500
Subtract y from both sides of the equation.
25\left(-y+2500\right)+20y=4800
Substitute -y+2500 for x in the other equation, 25x+20y=4800.
-25y+62500+20y=4800
Multiply 25 times -y+2500.
-5y+62500=4800
Add -25y to 20y.
-5y=-57700
Subtract 62500 from both sides of the equation.
y=11540
Divide both sides by -5.
x=-11540+2500
Substitute 11540 for y in x=-y+2500. Because the resulting equation contains only one variable, you can solve for x directly.
x=-9040
Add 2500 to -11540.
x=-9040,y=11540
The system is now solved.
x+y=2500,25x+20y=4800
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\25&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2500\\4800\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\25&20\end{matrix}\right))\left(\begin{matrix}1&1\\25&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\25&20\end{matrix}\right))\left(\begin{matrix}2500\\4800\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\25&20\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\\25&20\end{matrix}\right))\left(\begin{matrix}2500\\4800\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\\25&20\end{matrix}\right))\left(\begin{matrix}2500\\4800\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{20}{20-25}&-\frac{1}{20-25}\\-\frac{25}{20-25}&\frac{1}{20-25}\end{matrix}\right)\left(\begin{matrix}2500\\4800\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}{5}\\5&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}2500\\4800\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\times 2500+\frac{1}{5}\times 4800\\5\times 2500-\frac{1}{5}\times 4800\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-9040\\11540\end{matrix}\right)
Do the arithmetic.
x=-9040,y=11540
Extract the matrix elements x and y.
x+y=2500,25x+20y=4800
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.
25x+25y=25\times 2500,25x+20y=4800
To make x and 25x equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 1.
25x+25y=62500,25x+20y=4800
Simplify.
25x-25x+25y-20y=62500-4800
Subtract 25x+20y=4800 from 25x+25y=62500 by subtracting like terms on each side of the equal sign.
25y-20y=62500-4800
Add 25x to -25x. Terms 25x and -25x cancel out, leaving an equation with only one variable that can be solved.
5y=62500-4800
Add 25y to -20y.
5y=57700
Add 62500 to -4800.
y=11540
Divide both sides by 5.
25x+20\times 11540=4800
Substitute 11540 for y in 25x+20y=4800. Because the resulting equation contains only one variable, you can solve for x directly.
25x+230800=4800
Multiply 20 times 11540.
25x=-226000
Subtract 230800 from both sides of the equation.
x=-9040
Divide both sides by 25.
x=-9040,y=11540
The system is now solved.