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2x+4y=362,3x+2y=153.5
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+4y=362
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=-4y+362
Subtract 4y from both sides of the equation.
x=\frac{1}{2}\left(-4y+362\right)
Divide both sides by 2.
x=-2y+181
Multiply \frac{1}{2} times -4y+362.
3\left(-2y+181\right)+2y=153.5
Substitute -2y+181 for x in the other equation, 3x+2y=153.5.
-6y+543+2y=153.5
Multiply 3 times -2y+181.
-4y+543=153.5
Add -6y to 2y.
-4y=-389.5
Subtract 543 from both sides of the equation.
y=97.375
Divide both sides by -4.
x=-2\times 97.375+181
Substitute 97.375 for y in x=-2y+181. Because the resulting equation contains only one variable, you can solve for x directly.
x=-194.75+181
Multiply -2 times 97.375.
x=-13.75
Add 181 to -194.75.
x=-13.75,y=97.375
The system is now solved.
2x+4y=362,3x+2y=153.5
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&4\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}362\\153.5\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&4\\3&2\end{matrix}\right))\left(\begin{matrix}2&4\\3&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&4\\3&2\end{matrix}\right))\left(\begin{matrix}362\\153.5\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&4\\3&2\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&4\\3&2\end{matrix}\right))\left(\begin{matrix}362\\153.5\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&4\\3&2\end{matrix}\right))\left(\begin{matrix}362\\153.5\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{2}{2\times 2-4\times 3}&-\frac{4}{2\times 2-4\times 3}\\-\frac{3}{2\times 2-4\times 3}&\frac{2}{2\times 2-4\times 3}\end{matrix}\right)\left(\begin{matrix}362\\153.5\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}{4}&\frac{1}{2}\\\frac{3}{8}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}362\\153.5\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{4}\times 362+\frac{1}{2}\times 153.5\\\frac{3}{8}\times 362-\frac{1}{4}\times 153.5\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{55}{4}\\\frac{779}{8}\end{matrix}\right)
Do the arithmetic.
x=-\frac{55}{4},y=\frac{779}{8}
Extract the matrix elements x and y.
2x+4y=362,3x+2y=153.5
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.
3\times 2x+3\times 4y=3\times 362,2\times 3x+2\times 2y=2\times 153.5
To make 2x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 2.
6x+12y=1086,6x+4y=307
Simplify.
6x-6x+12y-4y=1086-307
Subtract 6x+4y=307 from 6x+12y=1086 by subtracting like terms on each side of the equal sign.
12y-4y=1086-307
Add 6x to -6x. Terms 6x and -6x cancel out, leaving an equation with only one variable that can be solved.
8y=1086-307
Add 12y to -4y.
8y=779
Add 1086 to -307.
y=\frac{779}{8}
Divide both sides by 8.
3x+2\times \frac{779}{8}=153.5
Substitute \frac{779}{8} for y in 3x+2y=153.5. Because the resulting equation contains only one variable, you can solve for x directly.
3x+\frac{779}{4}=153.5
Multiply 2 times \frac{779}{8}.
3x=-\frac{165}{4}
Subtract \frac{779}{4} from both sides of the equation.
x=-\frac{55}{4}
Divide both sides by 3.
x=-\frac{55}{4},y=\frac{779}{8}
The system is now solved.