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2n+y=200,2n-y=0
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.
2n+y=200
Choose one of the equations and solve it for n by isolating n on the left hand side of the equal sign.
2n=-y+200
Subtract y from both sides of the equation.
n=\frac{1}{2}\left(-y+200\right)
Divide both sides by 2.
n=-\frac{1}{2}y+100
Multiply \frac{1}{2} times -y+200.
2\left(-\frac{1}{2}y+100\right)-y=0
Substitute -\frac{y}{2}+100 for n in the other equation, 2n-y=0.
-y+200-y=0
Multiply 2 times -\frac{y}{2}+100.
-2y+200=0
Add -y to -y.
-2y=-200
Subtract 200 from both sides of the equation.
y=100
Divide both sides by -2.
n=-\frac{1}{2}\times 100+100
Substitute 100 for y in n=-\frac{1}{2}y+100. Because the resulting equation contains only one variable, you can solve for n directly.
n=-50+100
Multiply -\frac{1}{2} times 100.
n=50
Add 100 to -50.
n=50,y=100
The system is now solved.
2n+y=200,2n-y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}2&1\\2&-1\end{matrix}\right)\left(\begin{matrix}n\\y\end{matrix}\right)=\left(\begin{matrix}200\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}2&1\\2&-1\end{matrix}\right))\left(\begin{matrix}2&1\\2&-1\end{matrix}\right)\left(\begin{matrix}n\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\2&-1\end{matrix}\right))\left(\begin{matrix}200\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&1\\2&-1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}n\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\2&-1\end{matrix}\right))\left(\begin{matrix}200\\0\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}n\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&1\\2&-1\end{matrix}\right))\left(\begin{matrix}200\\0\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}n\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2\left(-1\right)-2}&-\frac{1}{2\left(-1\right)-2}\\-\frac{2}{2\left(-1\right)-2}&\frac{2}{2\left(-1\right)-2}\end{matrix}\right)\left(\begin{matrix}200\\0\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}n\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}&\frac{1}{4}\\\frac{1}{2}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}200\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}n\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{4}\times 200\\\frac{1}{2}\times 200\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}n\\y\end{matrix}\right)=\left(\begin{matrix}50\\100\end{matrix}\right)
Do the arithmetic.
n=50,y=100
Extract the matrix elements n and y.
2n+y=200,2n-y=0
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.
2n-2n+y+y=200
Subtract 2n-y=0 from 2n+y=200 by subtracting like terms on each side of the equal sign.
y+y=200
Add 2n to -2n. Terms 2n and -2n cancel out, leaving an equation with only one variable that can be solved.
2y=200
Add y to y.
y=100
Divide both sides by 2.
2n-100=0
Substitute 100 for y in 2n-y=0. Because the resulting equation contains only one variable, you can solve for n directly.
2n=100
Add 100 to both sides of the equation.
n=50
Divide both sides by 2.
n=50,y=100
The system is now solved.