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x+y=65,53x+50y=4100
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.
53\left(-y+65\right)+50y=4100
Substitute -y+65 for x in the other equation, 53x+50y=4100.
-53y+3445+50y=4100
Multiply 53 times -y+65.
-3y+3445=4100
Add -53y to 50y.
-3y=655
Subtract 3445 from both sides of the equation.
y=-\frac{655}{3}
Divide both sides by -3.
x=-\left(-\frac{655}{3}\right)+65
Substitute -\frac{655}{3} for y in x=-y+65. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{655}{3}+65
Multiply -1 times -\frac{655}{3}.
x=\frac{850}{3}
Add 65 to \frac{655}{3}.
x=\frac{850}{3},y=-\frac{655}{3}
The system is now solved.
x+y=65,53x+50y=4100
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\53&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}65\\4100\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\53&50\end{matrix}\right))\left(\begin{matrix}1&1\\53&50\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\53&50\end{matrix}\right))\left(\begin{matrix}65\\4100\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\53&50\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\\53&50\end{matrix}\right))\left(\begin{matrix}65\\4100\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\\53&50\end{matrix}\right))\left(\begin{matrix}65\\4100\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{50}{50-53}&-\frac{1}{50-53}\\-\frac{53}{50-53}&\frac{1}{50-53}\end{matrix}\right)\left(\begin{matrix}65\\4100\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{50}{3}&\frac{1}{3}\\\frac{53}{3}&-\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}65\\4100\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{50}{3}\times 65+\frac{1}{3}\times 4100\\\frac{53}{3}\times 65-\frac{1}{3}\times 4100\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{850}{3}\\-\frac{655}{3}\end{matrix}\right)
Do the arithmetic.
x=\frac{850}{3},y=-\frac{655}{3}
Extract the matrix elements x and y.
x+y=65,53x+50y=4100
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.
53x+53y=53\times 65,53x+50y=4100
To make x and 53x equal, multiply all terms on each side of the first equation by 53 and all terms on each side of the second by 1.
53x+53y=3445,53x+50y=4100
Simplify.
53x-53x+53y-50y=3445-4100
Subtract 53x+50y=4100 from 53x+53y=3445 by subtracting like terms on each side of the equal sign.
53y-50y=3445-4100
Add 53x to -53x. Terms 53x and -53x cancel out, leaving an equation with only one variable that can be solved.
3y=3445-4100
Add 53y to -50y.
3y=-655
Add 3445 to -4100.
y=-\frac{655}{3}
Divide both sides by 3.
53x+50\left(-\frac{655}{3}\right)=4100
Substitute -\frac{655}{3} for y in 53x+50y=4100. Because the resulting equation contains only one variable, you can solve for x directly.
53x-\frac{32750}{3}=4100
Multiply 50 times -\frac{655}{3}.
53x=\frac{45050}{3}
Add \frac{32750}{3} to both sides of the equation.
x=\frac{850}{3}
Divide both sides by 53.
x=\frac{850}{3},y=-\frac{655}{3}
The system is now solved.