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A-0.15B=90800
Consider the first equation. Subtract 0.15B from both sides.
B-0.2A=23600
Consider the second equation. Subtract 0.2A from both sides.
A-0.15B=90800,-0.2A+B=23600
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-0.15B=90800
Choose one of the equations and solve it for A by isolating A on the left hand side of the equal sign.
A=0.15B+90800
Add \frac{3B}{20} to both sides of the equation.
-0.2\left(0.15B+90800\right)+B=23600
Substitute \frac{3B}{20}+90800 for A in the other equation, -0.2A+B=23600.
-0.03B-18160+B=23600
Multiply -0.2 times \frac{3B}{20}+90800.
0.97B-18160=23600
Add -\frac{3B}{100} to B.
0.97B=41760
Add 18160 to both sides of the equation.
B=\frac{4176000}{97}
Divide both sides of the equation by 0.97, which is the same as multiplying both sides by the reciprocal of the fraction.
A=0.15\times \frac{4176000}{97}+90800
Substitute \frac{4176000}{97} for B in A=0.15B+90800. Because the resulting equation contains only one variable, you can solve for A directly.
A=\frac{626400}{97}+90800
Multiply 0.15 times \frac{4176000}{97} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
A=\frac{9434000}{97}
Add 90800 to \frac{626400}{97}.
A=\frac{9434000}{97},B=\frac{4176000}{97}
The system is now solved.
A-0.15B=90800
Consider the first equation. Subtract 0.15B from both sides.
B-0.2A=23600
Consider the second equation. Subtract 0.2A from both sides.
A-0.15B=90800,-0.2A+B=23600
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-0.15\\-0.2&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}90800\\23600\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-0.15\\-0.2&1\end{matrix}\right))\left(\begin{matrix}1&-0.15\\-0.2&1\end{matrix}\right)\left(\begin{matrix}A\\B\end{matrix}\right)=inverse(\left(\begin{matrix}1&-0.15\\-0.2&1\end{matrix}\right))\left(\begin{matrix}90800\\23600\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-0.15\\-0.2&1\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&-0.15\\-0.2&1\end{matrix}\right))\left(\begin{matrix}90800\\23600\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&-0.15\\-0.2&1\end{matrix}\right))\left(\begin{matrix}90800\\23600\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{1}{1-\left(-0.15\left(-0.2\right)\right)}&-\frac{-0.15}{1-\left(-0.15\left(-0.2\right)\right)}\\-\frac{-0.2}{1-\left(-0.15\left(-0.2\right)\right)}&\frac{1}{1-\left(-0.15\left(-0.2\right)\right)}\end{matrix}\right)\left(\begin{matrix}90800\\23600\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{100}{97}&\frac{15}{97}\\\frac{20}{97}&\frac{100}{97}\end{matrix}\right)\left(\begin{matrix}90800\\23600\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{100}{97}\times 90800+\frac{15}{97}\times 23600\\\frac{20}{97}\times 90800+\frac{100}{97}\times 23600\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}A\\B\end{matrix}\right)=\left(\begin{matrix}\frac{9434000}{97}\\\frac{4176000}{97}\end{matrix}\right)
Do the arithmetic.
A=\frac{9434000}{97},B=\frac{4176000}{97}
Extract the matrix elements A and B.
A-0.15B=90800
Consider the first equation. Subtract 0.15B from both sides.
B-0.2A=23600
Consider the second equation. Subtract 0.2A from both sides.
A-0.15B=90800,-0.2A+B=23600
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.
-0.2A-0.2\left(-0.15\right)B=-0.2\times 90800,-0.2A+B=23600
To make A and -\frac{A}{5} equal, multiply all terms on each side of the first equation by -0.2 and all terms on each side of the second by 1.
-0.2A+0.03B=-18160,-0.2A+B=23600
Simplify.
-0.2A+0.2A+0.03B-B=-18160-23600
Subtract -0.2A+B=23600 from -0.2A+0.03B=-18160 by subtracting like terms on each side of the equal sign.
0.03B-B=-18160-23600
Add -\frac{A}{5} to \frac{A}{5}. Terms -\frac{A}{5} and \frac{A}{5} cancel out, leaving an equation with only one variable that can be solved.
-0.97B=-18160-23600
Add \frac{3B}{100} to -B.
-0.97B=-41760
Add -18160 to -23600.
B=\frac{4176000}{97}
Divide both sides of the equation by -0.97, which is the same as multiplying both sides by the reciprocal of the fraction.
-0.2A+\frac{4176000}{97}=23600
Substitute \frac{4176000}{97} for B in -0.2A+B=23600. Because the resulting equation contains only one variable, you can solve for A directly.
-0.2A=-\frac{1886800}{97}
Subtract \frac{4176000}{97} from both sides of the equation.
A=\frac{9434000}{97}
Multiply both sides by -5.
A=\frac{9434000}{97},B=\frac{4176000}{97}
The system is now solved.