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a+b=120,30a+70b=8240
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.
a+b=120
Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.
a=-b+120
Subtract b from both sides of the equation.
30\left(-b+120\right)+70b=8240
Substitute -b+120 for a in the other equation, 30a+70b=8240.
-30b+3600+70b=8240
Multiply 30 times -b+120.
40b+3600=8240
Add -30b to 70b.
40b=4640
Subtract 3600 from both sides of the equation.
b=116
Divide both sides by 40.
a=-116+120
Substitute 116 for b in a=-b+120. Because the resulting equation contains only one variable, you can solve for a directly.
a=4
Add 120 to -116.
a=4,b=116
The system is now solved.
a+b=120,30a+70b=8240
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\30&70\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}120\\8240\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\30&70\end{matrix}\right))\left(\begin{matrix}1&1\\30&70\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&70\end{matrix}\right))\left(\begin{matrix}120\\8240\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\30&70\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&70\end{matrix}\right))\left(\begin{matrix}120\\8240\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&70\end{matrix}\right))\left(\begin{matrix}120\\8240\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{70}{70-30}&-\frac{1}{70-30}\\-\frac{30}{70-30}&\frac{1}{70-30}\end{matrix}\right)\left(\begin{matrix}120\\8240\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}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{7}{4}&-\frac{1}{40}\\-\frac{3}{4}&\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}120\\8240\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{7}{4}\times 120-\frac{1}{40}\times 8240\\-\frac{3}{4}\times 120+\frac{1}{40}\times 8240\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}4\\116\end{matrix}\right)
Do the arithmetic.
a=4,b=116
Extract the matrix elements a and b.
a+b=120,30a+70b=8240
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.
30a+30b=30\times 120,30a+70b=8240
To make a and 30a equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 1.
30a+30b=3600,30a+70b=8240
Simplify.
30a-30a+30b-70b=3600-8240
Subtract 30a+70b=8240 from 30a+30b=3600 by subtracting like terms on each side of the equal sign.
30b-70b=3600-8240
Add 30a to -30a. Terms 30a and -30a cancel out, leaving an equation with only one variable that can be solved.
-40b=3600-8240
Add 30b to -70b.
-40b=-4640
Add 3600 to -8240.
b=116
Divide both sides by -40.
30a+70\times 116=8240
Substitute 116 for b in 30a+70b=8240. Because the resulting equation contains only one variable, you can solve for a directly.
30a+8120=8240
Multiply 70 times 116.
30a=120
Subtract 8120 from both sides of the equation.
a=4
Divide both sides by 30.
a=4,b=116
The system is now solved.