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100x-800y=0
Consider the second equation. Subtract 800y from both sides.
x+y=80,100x-800y=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.
x+y=80
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+80
Subtract y from both sides of the equation.
100\left(-y+80\right)-800y=0
Substitute -y+80 for x in the other equation, 100x-800y=0.
-100y+8000-800y=0
Multiply 100 times -y+80.
-900y+8000=0
Add -100y to -800y.
-900y=-8000
Subtract 8000 from both sides of the equation.
y=\frac{80}{9}
Divide both sides by -900.
x=-\frac{80}{9}+80
Substitute \frac{80}{9} for y in x=-y+80. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{640}{9}
Add 80 to -\frac{80}{9}.
x=\frac{640}{9},y=\frac{80}{9}
The system is now solved.
100x-800y=0
Consider the second equation. Subtract 800y from both sides.
x+y=80,100x-800y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\100&-800\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}80\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\100&-800\end{matrix}\right))\left(\begin{matrix}1&1\\100&-800\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\100&-800\end{matrix}\right))\left(\begin{matrix}80\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\100&-800\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\\100&-800\end{matrix}\right))\left(\begin{matrix}80\\0\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\\100&-800\end{matrix}\right))\left(\begin{matrix}80\\0\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{800}{-800-100}&-\frac{1}{-800-100}\\-\frac{100}{-800-100}&\frac{1}{-800-100}\end{matrix}\right)\left(\begin{matrix}80\\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}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{9}&\frac{1}{900}\\\frac{1}{9}&-\frac{1}{900}\end{matrix}\right)\left(\begin{matrix}80\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{8}{9}\times 80\\\frac{1}{9}\times 80\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{640}{9}\\\frac{80}{9}\end{matrix}\right)
Do the arithmetic.
x=\frac{640}{9},y=\frac{80}{9}
Extract the matrix elements x and y.
100x-800y=0
Consider the second equation. Subtract 800y from both sides.
x+y=80,100x-800y=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.
100x+100y=100\times 80,100x-800y=0
To make x and 100x equal, multiply all terms on each side of the first equation by 100 and all terms on each side of the second by 1.
100x+100y=8000,100x-800y=0
Simplify.
100x-100x+100y+800y=8000
Subtract 100x-800y=0 from 100x+100y=8000 by subtracting like terms on each side of the equal sign.
100y+800y=8000
Add 100x to -100x. Terms 100x and -100x cancel out, leaving an equation with only one variable that can be solved.
900y=8000
Add 100y to 800y.
y=\frac{80}{9}
Divide both sides by 900.
100x-800\times \frac{80}{9}=0
Substitute \frac{80}{9} for y in 100x-800y=0. Because the resulting equation contains only one variable, you can solve for x directly.
100x-\frac{64000}{9}=0
Multiply -800 times \frac{80}{9}.
100x=\frac{64000}{9}
Add \frac{64000}{9} to both sides of the equation.
x=\frac{640}{9}
Divide both sides by 100.
x=\frac{640}{9},y=\frac{80}{9}
The system is now solved.