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x+10-y=0
Consider the first equation. Subtract y from both sides.
x-y=-10
Subtract 10 from both sides. Anything subtracted from zero gives its negation.
x-y=-10,x+y=992
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=-10
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=y-10
Add y to both sides of the equation.
y-10+y=992
Substitute y-10 for x in the other equation, x+y=992.
2y-10=992
Add y to y.
2y=1002
Add 10 to both sides of the equation.
y=501
Divide both sides by 2.
x=501-10
Substitute 501 for y in x=y-10. Because the resulting equation contains only one variable, you can solve for x directly.
x=491
Add -10 to 501.
x=491,y=501
The system is now solved.
x+10-y=0
Consider the first equation. Subtract y from both sides.
x-y=-10
Subtract 10 from both sides. Anything subtracted from zero gives its negation.
x-y=-10,x+y=992
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}-10\\992\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}-10\\992\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}-10\\992\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}-10\\992\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-\left(-1\right)}&-\frac{-1}{1-\left(-1\right)}\\-\frac{1}{1-\left(-1\right)}&\frac{1}{1-\left(-1\right)}\end{matrix}\right)\left(\begin{matrix}-10\\992\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}-10\\992\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\left(-10\right)+\frac{1}{2}\times 992\\-\frac{1}{2}\left(-10\right)+\frac{1}{2}\times 992\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}491\\501\end{matrix}\right)
Do the arithmetic.
x=491,y=501
Extract the matrix elements x and y.
x+10-y=0
Consider the first equation. Subtract y from both sides.
x-y=-10
Subtract 10 from both sides. Anything subtracted from zero gives its negation.
x-y=-10,x+y=992
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=-10-992
Subtract x+y=992 from x-y=-10 by subtracting like terms on each side of the equal sign.
-y-y=-10-992
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
-2y=-10-992
Add -y to -y.
-2y=-1002
Add -10 to -992.
y=501
Divide both sides by -2.
x+501=992
Substitute 501 for y in x+y=992. Because the resulting equation contains only one variable, you can solve for x directly.
x=491
Subtract 501 from both sides of the equation.
x=491,y=501
The system is now solved.