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x+y=120,2x+y=200
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=120
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+120
Subtract y from both sides of the equation.
2\left(-y+120\right)+y=200
Substitute -y+120 for x in the other equation, 2x+y=200.
-2y+240+y=200
Multiply 2 times -y+120.
-y+240=200
Add -2y to y.
-y=-40
Subtract 240 from both sides of the equation.
y=40
Divide both sides by -1.
x=-40+120
Substitute 40 for y in x=-y+120. Because the resulting equation contains only one variable, you can solve for x directly.
x=80
Add 120 to -40.
x=80,y=40
The system is now solved.
x+y=120,2x+y=200
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}120\\200\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\2&1\end{matrix}\right))\left(\begin{matrix}1&1\\2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&1\end{matrix}\right))\left(\begin{matrix}120\\200\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2&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\\2&1\end{matrix}\right))\left(\begin{matrix}120\\200\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\\2&1\end{matrix}\right))\left(\begin{matrix}120\\200\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}{1-2}&-\frac{1}{1-2}\\-\frac{2}{1-2}&\frac{1}{1-2}\end{matrix}\right)\left(\begin{matrix}120\\200\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}-1&1\\2&-1\end{matrix}\right)\left(\begin{matrix}120\\200\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-120+200\\2\times 120-200\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\40\end{matrix}\right)
Do the arithmetic.
x=80,y=40
Extract the matrix elements x and y.
x+y=120,2x+y=200
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-2x+y-y=120-200
Subtract 2x+y=200 from x+y=120 by subtracting like terms on each side of the equal sign.
x-2x=120-200
Add y to -y. Terms y and -y cancel out, leaving an equation with only one variable that can be solved.
-x=120-200
Add x to -2x.
-x=-80
Add 120 to -200.
x=80
Divide both sides by -1.
2\times 80+y=200
Substitute 80 for x in 2x+y=200. Because the resulting equation contains only one variable, you can solve for y directly.
160+y=200
Multiply 2 times 80.
y=40
Subtract 160 from both sides of the equation.
x=80,y=40
The system is now solved.