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10m+16n=140,5m-8n=60
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.
10m+16n=140
Choose one of the equations and solve it for m by isolating m on the left hand side of the equal sign.
10m=-16n+140
Subtract 16n from both sides of the equation.
m=\frac{1}{10}\left(-16n+140\right)
Divide both sides by 10.
m=-\frac{8}{5}n+14
Multiply \frac{1}{10} times -16n+140.
5\left(-\frac{8}{5}n+14\right)-8n=60
Substitute -\frac{8n}{5}+14 for m in the other equation, 5m-8n=60.
-8n+70-8n=60
Multiply 5 times -\frac{8n}{5}+14.
-16n+70=60
Add -8n to -8n.
-16n=-10
Subtract 70 from both sides of the equation.
n=\frac{5}{8}
Divide both sides by -16.
m=-\frac{8}{5}\times \frac{5}{8}+14
Substitute \frac{5}{8} for n in m=-\frac{8}{5}n+14. Because the resulting equation contains only one variable, you can solve for m directly.
m=-1+14
Multiply -\frac{8}{5} times \frac{5}{8} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
m=13
Add 14 to -1.
m=13,n=\frac{5}{8}
The system is now solved.
10m+16n=140,5m-8n=60
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}10&16\\5&-8\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}140\\60\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}10&16\\5&-8\end{matrix}\right))\left(\begin{matrix}10&16\\5&-8\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}10&16\\5&-8\end{matrix}\right))\left(\begin{matrix}140\\60\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&16\\5&-8\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}10&16\\5&-8\end{matrix}\right))\left(\begin{matrix}140\\60\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}m\\n\end{matrix}\right)=inverse(\left(\begin{matrix}10&16\\5&-8\end{matrix}\right))\left(\begin{matrix}140\\60\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}-\frac{8}{10\left(-8\right)-16\times 5}&-\frac{16}{10\left(-8\right)-16\times 5}\\-\frac{5}{10\left(-8\right)-16\times 5}&\frac{10}{10\left(-8\right)-16\times 5}\end{matrix}\right)\left(\begin{matrix}140\\60\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}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{1}{20}&\frac{1}{10}\\\frac{1}{32}&-\frac{1}{16}\end{matrix}\right)\left(\begin{matrix}140\\60\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}\frac{1}{20}\times 140+\frac{1}{10}\times 60\\\frac{1}{32}\times 140-\frac{1}{16}\times 60\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}m\\n\end{matrix}\right)=\left(\begin{matrix}13\\\frac{5}{8}\end{matrix}\right)
Do the arithmetic.
m=13,n=\frac{5}{8}
Extract the matrix elements m and n.
10m+16n=140,5m-8n=60
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.
5\times 10m+5\times 16n=5\times 140,10\times 5m+10\left(-8\right)n=10\times 60
To make 10m and 5m equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 10.
50m+80n=700,50m-80n=600
Simplify.
50m-50m+80n+80n=700-600
Subtract 50m-80n=600 from 50m+80n=700 by subtracting like terms on each side of the equal sign.
80n+80n=700-600
Add 50m to -50m. Terms 50m and -50m cancel out, leaving an equation with only one variable that can be solved.
160n=700-600
Add 80n to 80n.
160n=100
Add 700 to -600.
n=\frac{5}{8}
Divide both sides by 160.
5m-8\times \frac{5}{8}=60
Substitute \frac{5}{8} for n in 5m-8n=60. Because the resulting equation contains only one variable, you can solve for m directly.
5m-5=60
Multiply -8 times \frac{5}{8}.
5m=65
Add 5 to both sides of the equation.
m=13
Divide both sides by 5.
m=13,n=\frac{5}{8}
The system is now solved.