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110b+218c=-93,109b+227c=-99
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.
110b+218c=-93
Choose one of the equations and solve it for b by isolating b on the left hand side of the equal sign.
110b=-218c-93
Subtract 218c from both sides of the equation.
b=\frac{1}{110}\left(-218c-93\right)
Divide both sides by 110.
b=-\frac{109}{55}c-\frac{93}{110}
Multiply \frac{1}{110} times -218c-93.
109\left(-\frac{109}{55}c-\frac{93}{110}\right)+227c=-99
Substitute -\frac{109c}{55}-\frac{93}{110} for b in the other equation, 109b+227c=-99.
-\frac{11881}{55}c-\frac{10137}{110}+227c=-99
Multiply 109 times -\frac{109c}{55}-\frac{93}{110}.
\frac{604}{55}c-\frac{10137}{110}=-99
Add -\frac{11881c}{55} to 227c.
\frac{604}{55}c=-\frac{753}{110}
Add \frac{10137}{110} to both sides of the equation.
c=-\frac{753}{1208}
Divide both sides of the equation by \frac{604}{55}, which is the same as multiplying both sides by the reciprocal of the fraction.
b=-\frac{109}{55}\left(-\frac{753}{1208}\right)-\frac{93}{110}
Substitute -\frac{753}{1208} for c in b=-\frac{109}{55}c-\frac{93}{110}. Because the resulting equation contains only one variable, you can solve for b directly.
b=\frac{82077}{66440}-\frac{93}{110}
Multiply -\frac{109}{55} times -\frac{753}{1208} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
b=\frac{471}{1208}
Add -\frac{93}{110} to \frac{82077}{66440} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
b=\frac{471}{1208},c=-\frac{753}{1208}
The system is now solved.
110b+218c=-93,109b+227c=-99
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}110&218\\109&227\end{matrix}\right)\left(\begin{matrix}b\\c\end{matrix}\right)=\left(\begin{matrix}-93\\-99\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}110&218\\109&227\end{matrix}\right))\left(\begin{matrix}110&218\\109&227\end{matrix}\right)\left(\begin{matrix}b\\c\end{matrix}\right)=inverse(\left(\begin{matrix}110&218\\109&227\end{matrix}\right))\left(\begin{matrix}-93\\-99\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}110&218\\109&227\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\c\end{matrix}\right)=inverse(\left(\begin{matrix}110&218\\109&227\end{matrix}\right))\left(\begin{matrix}-93\\-99\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}b\\c\end{matrix}\right)=inverse(\left(\begin{matrix}110&218\\109&227\end{matrix}\right))\left(\begin{matrix}-93\\-99\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}b\\c\end{matrix}\right)=\left(\begin{matrix}\frac{227}{110\times 227-218\times 109}&-\frac{218}{110\times 227-218\times 109}\\-\frac{109}{110\times 227-218\times 109}&\frac{110}{110\times 227-218\times 109}\end{matrix}\right)\left(\begin{matrix}-93\\-99\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}b\\c\end{matrix}\right)=\left(\begin{matrix}\frac{227}{1208}&-\frac{109}{604}\\-\frac{109}{1208}&\frac{55}{604}\end{matrix}\right)\left(\begin{matrix}-93\\-99\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}b\\c\end{matrix}\right)=\left(\begin{matrix}\frac{227}{1208}\left(-93\right)-\frac{109}{604}\left(-99\right)\\-\frac{109}{1208}\left(-93\right)+\frac{55}{604}\left(-99\right)\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}b\\c\end{matrix}\right)=\left(\begin{matrix}\frac{471}{1208}\\-\frac{753}{1208}\end{matrix}\right)
Do the arithmetic.
b=\frac{471}{1208},c=-\frac{753}{1208}
Extract the matrix elements b and c.
110b+218c=-93,109b+227c=-99
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.
109\times 110b+109\times 218c=109\left(-93\right),110\times 109b+110\times 227c=110\left(-99\right)
To make 110b and 109b equal, multiply all terms on each side of the first equation by 109 and all terms on each side of the second by 110.
11990b+23762c=-10137,11990b+24970c=-10890
Simplify.
11990b-11990b+23762c-24970c=-10137+10890
Subtract 11990b+24970c=-10890 from 11990b+23762c=-10137 by subtracting like terms on each side of the equal sign.
23762c-24970c=-10137+10890
Add 11990b to -11990b. Terms 11990b and -11990b cancel out, leaving an equation with only one variable that can be solved.
-1208c=-10137+10890
Add 23762c to -24970c.
-1208c=753
Add -10137 to 10890.
c=-\frac{753}{1208}
Divide both sides by -1208.
109b+227\left(-\frac{753}{1208}\right)=-99
Substitute -\frac{753}{1208} for c in 109b+227c=-99. Because the resulting equation contains only one variable, you can solve for b directly.
109b-\frac{170931}{1208}=-99
Multiply 227 times -\frac{753}{1208}.
109b=\frac{51339}{1208}
Add \frac{170931}{1208} to both sides of the equation.
b=\frac{471}{1208}
Divide both sides by 109.
b=\frac{471}{1208},c=-\frac{753}{1208}
The system is now solved.