Solve for a, b
a=190
b=70
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a+b=260,40a+20b=9000
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=260
Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.
a=-b+260
Subtract b from both sides of the equation.
40\left(-b+260\right)+20b=9000
Substitute -b+260 for a in the other equation, 40a+20b=9000.
-40b+10400+20b=9000
Multiply 40 times -b+260.
-20b+10400=9000
Add -40b to 20b.
-20b=-1400
Subtract 10400 from both sides of the equation.
b=70
Divide both sides by -20.
a=-70+260
Substitute 70 for b in a=-b+260. Because the resulting equation contains only one variable, you can solve for a directly.
a=190
Add 260 to -70.
a=190,b=70
The system is now solved.
a+b=260,40a+20b=9000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\40&20\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}260\\9000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\40&20\end{matrix}\right))\left(\begin{matrix}1&1\\40&20\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\40&20\end{matrix}\right))\left(\begin{matrix}260\\9000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\40&20\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\\40&20\end{matrix}\right))\left(\begin{matrix}260\\9000\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\\40&20\end{matrix}\right))\left(\begin{matrix}260\\9000\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{20}{20-40}&-\frac{1}{20-40}\\-\frac{40}{20-40}&\frac{1}{20-40}\end{matrix}\right)\left(\begin{matrix}260\\9000\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}-1&\frac{1}{20}\\2&-\frac{1}{20}\end{matrix}\right)\left(\begin{matrix}260\\9000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-260+\frac{1}{20}\times 9000\\2\times 260-\frac{1}{20}\times 9000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}190\\70\end{matrix}\right)
Do the arithmetic.
a=190,b=70
Extract the matrix elements a and b.
a+b=260,40a+20b=9000
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.
40a+40b=40\times 260,40a+20b=9000
To make a and 40a equal, multiply all terms on each side of the first equation by 40 and all terms on each side of the second by 1.
40a+40b=10400,40a+20b=9000
Simplify.
40a-40a+40b-20b=10400-9000
Subtract 40a+20b=9000 from 40a+40b=10400 by subtracting like terms on each side of the equal sign.
40b-20b=10400-9000
Add 40a to -40a. Terms 40a and -40a cancel out, leaving an equation with only one variable that can be solved.
20b=10400-9000
Add 40b to -20b.
20b=1400
Add 10400 to -9000.
b=70
Divide both sides by 20.
40a+20\times 70=9000
Substitute 70 for b in 40a+20b=9000. Because the resulting equation contains only one variable, you can solve for a directly.
40a+1400=9000
Multiply 20 times 70.
40a=7600
Subtract 1400 from both sides of the equation.
a=190
Divide both sides by 40.
a=190,b=70
The system is now solved.
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Simultaneous equation
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Limits
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