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3A+2B=120,5A+4B=210
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.
3A+2B=120
Choose one of the equations and solve it for A by isolating A on the left hand side of the equal sign.
3A=-2B+120
Subtract 2B from both sides of the equation.
A=\frac{1}{3}\left(-2B+120\right)
Divide both sides by 3.
A=-\frac{2}{3}B+40
Multiply \frac{1}{3} times -2B+120.
5\left(-\frac{2}{3}B+40\right)+4B=210
Substitute -\frac{2B}{3}+40 for A in the other equation, 5A+4B=210.
-\frac{10}{3}B+200+4B=210
Multiply 5 times -\frac{2B}{3}+40.
\frac{2}{3}B+200=210
Add -\frac{10B}{3} to 4B.
\frac{2}{3}B=10
Subtract 200 from both sides of the equation.
B=15
Divide both sides of the equation by \frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
A=-\frac{2}{3}\times 15+40
Substitute 15 for B in A=-\frac{2}{3}B+40. Because the resulting equation contains only one variable, you can solve for A directly.
A=-10+40
Multiply -\frac{2}{3} times 15.
A=30
Add 40 to -10.
A=30,B=15
The system is now solved.
3A+2B=120,5A+4B=210
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&2\\5&4\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}120\\210\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&2\\5&4\end{matrix}\right))\left(\begin{matrix}3&2\\5&4\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\5&4\end{matrix}\right))\left(\begin{matrix}120\\210\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\5&4\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}3&2\\5&4\end{matrix}\right))\left(\begin{matrix}120\\210\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}3&2\\5&4\end{matrix}\right))\left(\begin{matrix}120\\210\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{4}{3\times 4-2\times 5}&-\frac{2}{3\times 4-2\times 5}\\-\frac{5}{3\times 4-2\times 5}&\frac{3}{3\times 4-2\times 5}\end{matrix}\right)\left(\begin{matrix}120\\210\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}2&-1\\-\frac{5}{2}&\frac{3}{2}\end{matrix}\right)\left(\begin{matrix}120\\210\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}2\times 120-210\\-\frac{5}{2}\times 120+\frac{3}{2}\times 210\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}30\\15\end{matrix}\right)
Do the arithmetic.
A=30,B=15
Extract the matrix elements A and B.
3A+2B=120,5A+4B=210
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 3A+5\times 2B=5\times 120,3\times 5A+3\times 4B=3\times 210
To make 3A and 5A equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 3.
15A+10B=600,15A+12B=630
Simplify.
15A-15A+10B-12B=600-630
Subtract 15A+12B=630 from 15A+10B=600 by subtracting like terms on each side of the equal sign.
10B-12B=600-630
Add 15A to -15A. Terms 15A and -15A cancel out, leaving an equation with only one variable that can be solved.
-2B=600-630
Add 10B to -12B.
-2B=-30
Add 600 to -630.
B=15
Divide both sides by -2.
5A+4\times 15=210
Substitute 15 for B in 5A+4B=210. Because the resulting equation contains only one variable, you can solve for A directly.
5A+60=210
Multiply 4 times 15.
5A=150
Subtract 60 from both sides of the equation.
A=30
Divide both sides by 5.
A=30,B=15
The system is now solved.