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x+y=65,40x+20y=2100
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=65
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+65
Subtract y from both sides of the equation.
40\left(-y+65\right)+20y=2100
Substitute -y+65 for x in the other equation, 40x+20y=2100.
-40y+2600+20y=2100
Multiply 40 times -y+65.
-20y+2600=2100
Add -40y to 20y.
-20y=-500
Subtract 2600 from both sides of the equation.
y=25
Divide both sides by -20.
x=-25+65
Substitute 25 for y in x=-y+65. Because the resulting equation contains only one variable, you can solve for x directly.
x=40
Add 65 to -25.
x=40,y=25
The system is now solved.
x+y=65,40x+20y=2100
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\40&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}65\\2100\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\40&20\end{matrix}\right))\left(\begin{matrix}1&1\\40&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\40&20\end{matrix}\right))\left(\begin{matrix}65\\2100\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\40&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\\40&20\end{matrix}\right))\left(\begin{matrix}65\\2100\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\\40&20\end{matrix}\right))\left(\begin{matrix}65\\2100\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}{20-40}&-\frac{1}{20-40}\\-\frac{40}{20-40}&\frac{1}{20-40}\end{matrix}\right)\left(\begin{matrix}65\\2100\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}{20}\\2&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}65\\2100\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-65+\frac{1}{20}\times 2100\\2\times 65-\frac{1}{20}\times 2100\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\25\end{matrix}\right)
Do the arithmetic.
x=40,y=25
Extract the matrix elements x and y.
x+y=65,40x+20y=2100
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.
40x+40y=40\times 65,40x+20y=2100
To make x and 40x equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 1.
40x+40y=2600,40x+20y=2100
Simplify.
40x-40x+40y-20y=2600-2100
Subtract 40x+20y=2100 from 40x+40y=2600 by subtracting like terms on each side of the equal sign.
40y-20y=2600-2100
Add 40x to -40x. Terms 40x and -40x cancel out, leaving an equation with only one variable that can be solved.
20y=2600-2100
Add 40y to -20y.
20y=500
Add 2600 to -2100.
y=25
Divide both sides by 20.
40x+20\times 25=2100
Substitute 25 for y in 40x+20y=2100. Because the resulting equation contains only one variable, you can solve for x directly.
40x+500=2100
Multiply 20 times 25.
40x=1600
Subtract 500 from both sides of the equation.
x=40
Divide both sides by 40.
x=40,y=25
The system is now solved.