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x+200y=1800,x+180y=1700
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+200y=1800
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-200y+1800
Subtract 200y from both sides of the equation.
-200y+1800+180y=1700
Substitute -200y+1800 for x in the other equation, x+180y=1700.
-20y+1800=1700
Add -200y to 180y.
-20y=-100
Subtract 1800 from both sides of the equation.
y=5
Divide both sides by -20.
x=-200\times 5+1800
Substitute 5 for y in x=-200y+1800. Because the resulting equation contains only one variable, you can solve for x directly.
x=-1000+1800
Multiply -200 times 5.
x=800
Add 1800 to -1000.
x=800,y=5
The system is now solved.
x+200y=1800,x+180y=1700
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&200\\1&180\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1800\\1700\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&200\\1&180\end{matrix}\right))\left(\begin{matrix}1&200\\1&180\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&200\\1&180\end{matrix}\right))\left(\begin{matrix}1800\\1700\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&200\\1&180\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&200\\1&180\end{matrix}\right))\left(\begin{matrix}1800\\1700\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&200\\1&180\end{matrix}\right))\left(\begin{matrix}1800\\1700\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{180}{180-200}&-\frac{200}{180-200}\\-\frac{1}{180-200}&\frac{1}{180-200}\end{matrix}\right)\left(\begin{matrix}1800\\1700\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}-9&10\\\frac{1}{20}&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}1800\\1700\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-9\times 1800+10\times 1700\\\frac{1}{20}\times 1800-\frac{1}{20}\times 1700\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}800\\5\end{matrix}\right)
Do the arithmetic.
x=800,y=5
Extract the matrix elements x and y.
x+200y=1800,x+180y=1700
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+200y-180y=1800-1700
Subtract x+180y=1700 from x+200y=1800 by subtracting like terms on each side of the equal sign.
200y-180y=1800-1700
Add x to -x. Terms x and -x cancel out, leaving an equation with only one variable that can be solved.
20y=1800-1700
Add 200y to -180y.
20y=100
Add 1800 to -1700.
y=5
Divide both sides by 20.
x+180\times 5=1700
Substitute 5 for y in x+180y=1700. Because the resulting equation contains only one variable, you can solve for x directly.
x+900=1700
Multiply 180 times 5.
x=800
Subtract 900 from both sides of the equation.
x=800,y=5
The system is now solved.