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x+y=27,x-y=-145
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=27
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+27
Subtract y from both sides of the equation.
-y+27-y=-145
Substitute -y+27 for x in the other equation, x-y=-145.
-2y+27=-145
Add -y to -y.
-2y=-172
Subtract 27 from both sides of the equation.
y=86
Divide both sides by -2.
x=-86+27
Substitute 86 for y in x=-y+27. Because the resulting equation contains only one variable, you can solve for x directly.
x=-59
Add 27 to -86.
x=-59,y=86
The system is now solved.
x+y=27,x-y=-145
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}27\\-145\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}1&1\\1&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}27\\-145\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1&-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}1&1\\1&-1\end{matrix}\right))\left(\begin{matrix}27\\-145\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\\1&-1\end{matrix}\right))\left(\begin{matrix}27\\-145\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}{-1-1}&-\frac{1}{-1-1}\\-\frac{1}{-1-1}&\frac{1}{-1-1}\end{matrix}\right)\left(\begin{matrix}27\\-145\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}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}27\\-145\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 27+\frac{1}{2}\left(-145\right)\\\frac{1}{2}\times 27-\frac{1}{2}\left(-145\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-59\\86\end{matrix}\right)
Do the arithmetic.
x=-59,y=86
Extract the matrix elements x and y.
x+y=27,x-y=-145
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.
x-x+y+y=27+145
Subtract x-y=-145 from x+y=27 by subtracting like terms on each side of the equal sign.
y+y=27+145
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
2y=27+145
Add y to y.
2y=172
Add 27 to 145.
y=86
Divide both sides by 2.
x-86=-145
Substitute 86 for y in x-y=-145. Because the resulting equation contains only one variable, you can solve for x directly.
x=-59
Add 86 to both sides of the equation.
x=-59,y=86
The system is now solved.