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2x-y=16,2x+y=180
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-y=16
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
2x=y+16
Add y to both sides of the equation.
x=\frac{1}{2}\left(y+16\right)
Divide both sides by 2.
x=\frac{1}{2}y+8
Multiply \frac{1}{2} times y+16.
2\left(\frac{1}{2}y+8\right)+y=180
Substitute \frac{y}{2}+8 for x in the other equation, 2x+y=180.
y+16+y=180
Multiply 2 times \frac{y}{2}+8.
2y+16=180
Add y to y.
2y=164
Subtract 16 from both sides of the equation.
y=82
Divide both sides by 2.
x=\frac{1}{2}\times 82+8
Substitute 82 for y in x=\frac{1}{2}y+8. Because the resulting equation contains only one variable, you can solve for x directly.
x=41+8
Multiply \frac{1}{2} times 82.
x=49
Add 8 to 41.
x=49,y=82
The system is now solved.
2x-y=16,2x+y=180
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&-1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}16\\180\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&-1\\2&1\end{matrix}\right))\left(\begin{matrix}2&-1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-1\\2&1\end{matrix}\right))\left(\begin{matrix}16\\180\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-1\\2&1\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&-1\\2&1\end{matrix}\right))\left(\begin{matrix}16\\180\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&-1\\2&1\end{matrix}\right))\left(\begin{matrix}16\\180\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{1}{2-\left(-2\right)}&-\frac{-1}{2-\left(-2\right)}\\-\frac{2}{2-\left(-2\right)}&\frac{2}{2-\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}16\\180\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}{4}\\-\frac{1}{2}&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}16\\180\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\times 16+\frac{1}{4}\times 180\\-\frac{1}{2}\times 16+\frac{1}{2}\times 180\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}49\\82\end{matrix}\right)
Do the arithmetic.
x=49,y=82
Extract the matrix elements x and y.
2x-y=16,2x+y=180
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.
2x-2x-y-y=16-180
Subtract 2x+y=180 from 2x-y=16 by subtracting like terms on each side of the equal sign.
-y-y=16-180
Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.
-2y=16-180
Add -y to -y.
-2y=-164
Add 16 to -180.
y=82
Divide both sides by -2.
2x+82=180
Substitute 82 for y in 2x+y=180. Because the resulting equation contains only one variable, you can solve for x directly.
2x=98
Subtract 82 from both sides of the equation.
x=49
Divide both sides by 2.
x=49,y=82
The system is now solved.