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x+y=1,x-y=999
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=1
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+1
Subtract y from both sides of the equation.
-y+1-y=999
Substitute -y+1 for x in the other equation, x-y=999.
-2y+1=999
Add -y to -y.
-2y=998
Subtract 1 from both sides of the equation.
y=-499
Divide both sides by -2.
x=-\left(-499\right)+1
Substitute -499 for y in x=-y+1. Because the resulting equation contains only one variable, you can solve for x directly.
x=499+1
Multiply -1 times -499.
x=500
Add 1 to 499.
x=500,y=-499
The system is now solved.
x+y=1,x-y=999
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}1\\999\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}1\\999\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}1\\999\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}1\\999\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}1\\999\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}1\\999\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}+\frac{1}{2}\times 999\\\frac{1}{2}-\frac{1}{2}\times 999\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}500\\-499\end{matrix}\right)
Do the arithmetic.
x=500,y=-499
Extract the matrix elements x and y.
x+y=1,x-y=999
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=1-999
Subtract x-y=999 from x+y=1 by subtracting like terms on each side of the equal sign.
y+y=1-999
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
2y=1-999
Add y to y.
2y=-998
Add 1 to -999.
y=-499
Divide both sides by 2.
x-\left(-499\right)=999
Substitute -499 for y in x-y=999. Because the resulting equation contains only one variable, you can solve for x directly.
x+499=999
Multiply -1 times -499.
x=500
Subtract 499 from both sides of the equation.
x=500,y=-499
The system is now solved.