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2x+12y=275,5x+3y=242
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.
2x+12y=275
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-12y+275
Subtract 12y from both sides of the equation.
x=\frac{1}{2}\left(-12y+275\right)
Divide both sides by 2.
x=-6y+\frac{275}{2}
Multiply \frac{1}{2} times -12y+275.
5\left(-6y+\frac{275}{2}\right)+3y=242
Substitute -6y+\frac{275}{2} for x in the other equation, 5x+3y=242.
-30y+\frac{1375}{2}+3y=242
Multiply 5 times -6y+\frac{275}{2}.
-27y+\frac{1375}{2}=242
Add -30y to 3y.
-27y=-\frac{891}{2}
Subtract \frac{1375}{2} from both sides of the equation.
y=\frac{33}{2}
Divide both sides by -27.
x=-6\times \frac{33}{2}+\frac{275}{2}
Substitute \frac{33}{2} for y in x=-6y+\frac{275}{2}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-99+\frac{275}{2}
Multiply -6 times \frac{33}{2}.
x=\frac{77}{2}
Add \frac{275}{2} to -99.
x=\frac{77}{2},y=\frac{33}{2}
The system is now solved.
2x+12y=275,5x+3y=242
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&12\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}275\\242\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&12\\5&3\end{matrix}\right))\left(\begin{matrix}2&12\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&12\\5&3\end{matrix}\right))\left(\begin{matrix}275\\242\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&12\\5&3\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}2&12\\5&3\end{matrix}\right))\left(\begin{matrix}275\\242\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}2&12\\5&3\end{matrix}\right))\left(\begin{matrix}275\\242\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{3}{2\times 3-12\times 5}&-\frac{12}{2\times 3-12\times 5}\\-\frac{5}{2\times 3-12\times 5}&\frac{2}{2\times 3-12\times 5}\end{matrix}\right)\left(\begin{matrix}275\\242\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}{18}&\frac{2}{9}\\\frac{5}{54}&-\frac{1}{27}\end{matrix}\right)\left(\begin{matrix}275\\242\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{18}\times 275+\frac{2}{9}\times 242\\\frac{5}{54}\times 275-\frac{1}{27}\times 242\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{77}{2}\\\frac{33}{2}\end{matrix}\right)
Do the arithmetic.
x=\frac{77}{2},y=\frac{33}{2}
Extract the matrix elements x and y.
2x+12y=275,5x+3y=242
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.
5\times 2x+5\times 12y=5\times 275,2\times 5x+2\times 3y=2\times 242
To make 2x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 2.
10x+60y=1375,10x+6y=484
Simplify.
10x-10x+60y-6y=1375-484
Subtract 10x+6y=484 from 10x+60y=1375 by subtracting like terms on each side of the equal sign.
60y-6y=1375-484
Add 10x to -10x. Terms 10x and -10x cancel out, leaving an equation with only one variable that can be solved.
54y=1375-484
Add 60y to -6y.
54y=891
Add 1375 to -484.
y=\frac{33}{2}
Divide both sides by 54.
5x+3\times \frac{33}{2}=242
Substitute \frac{33}{2} for y in 5x+3y=242. Because the resulting equation contains only one variable, you can solve for x directly.
5x+\frac{99}{2}=242
Multiply 3 times \frac{33}{2}.
5x=\frac{385}{2}
Subtract \frac{99}{2} from both sides of the equation.
x=\frac{77}{2}
Divide both sides by 5.
x=\frac{77}{2},y=\frac{33}{2}
The system is now solved.