Solve {l}{63b+74d=11,133}{55b+72d=10,385}

$$\left. \begin{array} { l } { 63 b + 74 d = 11,133 } \\ { 55 b + 72 d = 10,385 } \end{array} \right.$$

Solve for b, d

$$b=71$$

$$d=90$$

Steps Using Substitution

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.

$$63b+74d=11133,55b+72d=10385$$

Choose one of the equations and solve it for b by isolating b on the left hand side of the equal sign.

$$63b+74d=11133$$

Subtract 74d from both sides of the equation.

$$63b=-74d+11133$$

Divide both sides by 63.

$$b=\frac{1}{63}\left(-74d+11133\right)$$

Multiply \frac{1}{63} times -74d+11133.

$$b=-\frac{74}{63}d+\frac{1237}{7}$$

Substitute -\frac{74d}{63}+\frac{1237}{7} for b in the other equation, 55b+72d=10385.

$$55\left(-\frac{74}{63}d+\frac{1237}{7}\right)+72d=10385$$

Multiply 55 times -\frac{74d}{63}+\frac{1237}{7}.

$$-\frac{4070}{63}d+\frac{68035}{7}+72d=10385$$

Add -\frac{4070d}{63} to 72d.

$$\frac{466}{63}d+\frac{68035}{7}=10385$$

Subtract \frac{68035}{7} from both sides of the equation.

$$\frac{466}{63}d=\frac{4660}{7}$$

Divide both sides of the equation by \frac{466}{63}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$d=90$$

Substitute 90 for d in b=-\frac{74}{63}d+\frac{1237}{7}. Because the resulting equation contains only one variable, you can solve for b directly.

$$b=-\frac{74}{63}\times 90+\frac{1237}{7}$$

Multiply -\frac{74}{63} times 90.

$$b=\frac{-740+1237}{7}$$

Add \frac{1237}{7} to -\frac{740}{7} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$b=71$$

The system is now solved.

$$b=71,d=90$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$63b+74d=11133,55b+72d=10385$$

Write the equations in matrix form.

$$\left(\begin{matrix}63&74\\55&72\end{matrix}\right)\left(\begin{matrix}b\\d\end{matrix}\right)=\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}63&74\\55&72\end{matrix}\right).

$$inverse(\left(\begin{matrix}63&74\\55&72\end{matrix}\right))\left(\begin{matrix}63&74\\55&72\end{matrix}\right)\left(\begin{matrix}b\\d\end{matrix}\right)=inverse(\left(\begin{matrix}63&74\\55&72\end{matrix}\right))\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\d\end{matrix}\right)=inverse(\left(\begin{matrix}63&74\\55&72\end{matrix}\right))\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}b\\d\end{matrix}\right)=inverse(\left(\begin{matrix}63&74\\55&72\end{matrix}\right))\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

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.

$$\left(\begin{matrix}b\\d\end{matrix}\right)=\left(\begin{matrix}\frac{72}{63\times 72-74\times 55}&-\frac{74}{63\times 72-74\times 55}\\-\frac{55}{63\times 72-74\times 55}&\frac{63}{63\times 72-74\times 55}\end{matrix}\right)\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\d\end{matrix}\right)=\left(\begin{matrix}\frac{36}{233}&-\frac{37}{233}\\-\frac{55}{466}&\frac{63}{466}\end{matrix}\right)\left(\begin{matrix}11133\\10385\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}b\\d\end{matrix}\right)=\left(\begin{matrix}\frac{36}{233}\times 11133-\frac{37}{233}\times 10385\\-\frac{55}{466}\times 11133+\frac{63}{466}\times 10385\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}b\\d\end{matrix}\right)=\left(\begin{matrix}71\\90\end{matrix}\right)$$

Extract the matrix elements b and d.

$$b=71,d=90$$

Steps Using Elimination

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.

$$63b+74d=11133,55b+72d=10385$$

To make 63b and 55b equal, multiply all terms on each side of the first equation by 55 and all terms on each side of the second by 63.

$$55\times 63b+55\times 74d=55\times 11133,63\times 55b+63\times 72d=63\times 10385$$

Simplify.

$$3465b+4070d=612315,3465b+4536d=654255$$

Subtract 3465b+4536d=654255 from 3465b+4070d=612315 by subtracting like terms on each side of the equal sign.

$$3465b-3465b+4070d-4536d=612315-654255$$

Add 3465b to -3465b. Terms 3465b and -3465b cancel out, leaving an equation with only one variable that can be solved.

$$4070d-4536d=612315-654255$$

Add 4070d to -4536d.

$$-466d=612315-654255$$

Add 612315 to -654255.

$$-466d=-41940$$

Divide both sides by -466.

$$d=90$$

Substitute 90 for d in 55b+72d=10385. Because the resulting equation contains only one variable, you can solve for b directly.

$$55b+72\times 90=10385$$

Multiply 72 times 90.

$$55b+6480=10385$$

Subtract 6480 from both sides of the equation.

$$55b=3905$$

Divide both sides by 55.

$$b=71$$

The system is now solved.

$$b=71,d=90$$