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x+y=92,50x+5y=1000
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=92
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+92
Subtract y from both sides of the equation.
50\left(-y+92\right)+5y=1000
Substitute -y+92 for x in the other equation, 50x+5y=1000.
-50y+4600+5y=1000
Multiply 50 times -y+92.
-45y+4600=1000
Add -50y to 5y.
-45y=-3600
Subtract 4600 from both sides of the equation.
y=80
Divide both sides by -45.
x=-80+92
Substitute 80 for y in x=-y+92. Because the resulting equation contains only one variable, you can solve for x directly.
x=12
Add 92 to -80.
x=12,y=80
The system is now solved.
x+y=92,50x+5y=1000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\50&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}92\\1000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\50&5\end{matrix}\right))\left(\begin{matrix}1&1\\50&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\50&5\end{matrix}\right))\left(\begin{matrix}92\\1000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\50&5\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\\50&5\end{matrix}\right))\left(\begin{matrix}92\\1000\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\\50&5\end{matrix}\right))\left(\begin{matrix}92\\1000\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{5}{5-50}&-\frac{1}{5-50}\\-\frac{50}{5-50}&\frac{1}{5-50}\end{matrix}\right)\left(\begin{matrix}92\\1000\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}{9}&\frac{1}{45}\\\frac{10}{9}&-\frac{1}{45}\end{matrix}\right)\left(\begin{matrix}92\\1000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{9}\times 92+\frac{1}{45}\times 1000\\\frac{10}{9}\times 92-\frac{1}{45}\times 1000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\80\end{matrix}\right)
Do the arithmetic.
x=12,y=80
Extract the matrix elements x and y.
x+y=92,50x+5y=1000
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.
50x+50y=50\times 92,50x+5y=1000
To make x and 50x equal, multiply all terms on each side of the first equation by 50 and all terms on each side of the second by 1.
50x+50y=4600,50x+5y=1000
Simplify.
50x-50x+50y-5y=4600-1000
Subtract 50x+5y=1000 from 50x+50y=4600 by subtracting like terms on each side of the equal sign.
50y-5y=4600-1000
Add 50x to -50x. Terms 50x and -50x cancel out, leaving an equation with only one variable that can be solved.
45y=4600-1000
Add 50y to -5y.
45y=3600
Add 4600 to -1000.
y=80
Divide both sides by 45.
50x+5\times 80=1000
Substitute 80 for y in 50x+5y=1000. Because the resulting equation contains only one variable, you can solve for x directly.
50x+400=1000
Multiply 5 times 80.
50x=600
Subtract 400 from both sides of the equation.
x=12
Divide both sides by 50.
x=12,y=80
The system is now solved.