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x+y=111,x-y=101
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=111
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+111
Subtract y from both sides of the equation.
-y+111-y=101
Substitute -y+111 for x in the other equation, x-y=101.
-2y+111=101
Add -y to -y.
-2y=-10
Subtract 111 from both sides of the equation.
y=5
Divide both sides by -2.
x=-5+111
Substitute 5 for y in x=-y+111. Because the resulting equation contains only one variable, you can solve for x directly.
x=106
Add 111 to -5.
x=106,y=5
The system is now solved.
x+y=111,x-y=101
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}111\\101\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}111\\101\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}111\\101\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}111\\101\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}111\\101\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}111\\101\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 111+\frac{1}{2}\times 101\\\frac{1}{2}\times 111-\frac{1}{2}\times 101\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}106\\5\end{matrix}\right)
Do the arithmetic.
x=106,y=5
Extract the matrix elements x and y.
x+y=111,x-y=101
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=111-101
Subtract x-y=101 from x+y=111 by subtracting like terms on each side of the equal sign.
y+y=111-101
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
2y=111-101
Add y to y.
2y=10
Add 111 to -101.
y=5
Divide both sides by 2.
x-5=101
Substitute 5 for y in x-y=101. Because the resulting equation contains only one variable, you can solve for x directly.
x=106
Add 5 to both sides of the equation.
x=106,y=5
The system is now solved.