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y-20-x=0
Consider the second equation. Subtract x from both sides.
y-x=20
Add 20 to both sides. Anything plus zero gives itself.
x+y=200,-x+y=20
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=200
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+200
Subtract y from both sides of the equation.
-\left(-y+200\right)+y=20
Substitute -y+200 for x in the other equation, -x+y=20.
y-200+y=20
Multiply -1 times -y+200.
2y-200=20
Add y to y.
2y=220
Add 200 to both sides of the equation.
y=110
Divide both sides by 2.
x=-110+200
Substitute 110 for y in x=-y+200. Because the resulting equation contains only one variable, you can solve for x directly.
x=90
Add 200 to -110.
x=90,y=110
The system is now solved.
y-20-x=0
Consider the second equation. Subtract x from both sides.
y-x=20
Add 20 to both sides. Anything plus zero gives itself.
x+y=200,-x+y=20
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}200\\20\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}200\\20\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}200\\20\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}200\\20\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}200\\20\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}200\\20\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 200-\frac{1}{2}\times 20\\\frac{1}{2}\times 200+\frac{1}{2}\times 20\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}90\\110\end{matrix}\right)
Do the arithmetic.
x=90,y=110
Extract the matrix elements x and y.
y-20-x=0
Consider the second equation. Subtract x from both sides.
y-x=20
Add 20 to both sides. Anything plus zero gives itself.
x+y=200,-x+y=20
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=200-20
Subtract -x+y=20 from x+y=200 by subtracting like terms on each side of the equal sign.
x+x=200-20
Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.
2x=200-20
Add x to x.
2x=180
Add 200 to -20.
x=90
Divide both sides by 2.
-90+y=20
Substitute 90 for x in -x+y=20. Because the resulting equation contains only one variable, you can solve for y directly.
y=110
Add 90 to both sides of the equation.
x=90,y=110
The system is now solved.