Solve {l}{30a+44b=10}{40a+55b=13}

$$\left. \begin{array} { l } { 30 a + 44 b = 10 } \\ { 40 a + 55 b = 13 } \end{array} \right.$$

Solve for a, b

$$a=\frac{1}{5}=0.2$$

$$b=\frac{1}{11}\approx 0.090909091$$

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.

$$30a+44b=10,40a+55b=13$$

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

$$30a+44b=10$$

Subtract 44b from both sides of the equation.

$$30a=-44b+10$$

Divide both sides by 30.

$$a=\frac{1}{30}\left(-44b+10\right)$$

Multiply \frac{1}{30} times -44b+10.

$$a=-\frac{22}{15}b+\frac{1}{3}$$

Substitute -\frac{22b}{15}+\frac{1}{3} for a in the other equation, 40a+55b=13.

$$40\left(-\frac{22}{15}b+\frac{1}{3}\right)+55b=13$$

Multiply 40 times -\frac{22b}{15}+\frac{1}{3}.

$$-\frac{176}{3}b+\frac{40}{3}+55b=13$$

Add -\frac{176b}{3} to 55b.

$$-\frac{11}{3}b+\frac{40}{3}=13$$

Subtract \frac{40}{3} from both sides of the equation.

$$-\frac{11}{3}b=-\frac{1}{3}$$

Divide both sides of the equation by -\frac{11}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{1}{11}$$

Substitute \frac{1}{11} for b in a=-\frac{22}{15}b+\frac{1}{3}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{22}{15}\times \frac{1}{11}+\frac{1}{3}$$

Multiply -\frac{22}{15} times \frac{1}{11} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{2}{15}+\frac{1}{3}$$

Add \frac{1}{3} to -\frac{2}{15} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{1}{5}$$

The system is now solved.

$$a=\frac{1}{5},b=\frac{1}{11}$$

Steps Using Matrices

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

$$30a+44b=10,40a+55b=13$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&44\\40&55\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}10\\13\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&44\\40&55\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&44\\40&55\end{matrix}\right))\left(\begin{matrix}30&44\\40&55\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}30&44\\40&55\end{matrix}\right))\left(\begin{matrix}10\\13\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}30&44\\40&55\end{matrix}\right))\left(\begin{matrix}10\\13\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}30&44\\40&55\end{matrix}\right))\left(\begin{matrix}10\\13\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{55}{30\times 55-44\times 40}&-\frac{44}{30\times 55-44\times 40}\\-\frac{40}{30\times 55-44\times 40}&\frac{30}{30\times 55-44\times 40}\end{matrix}\right)\left(\begin{matrix}10\\13\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}&\frac{2}{5}\\\frac{4}{11}&-\frac{3}{11}\end{matrix}\right)\left(\begin{matrix}10\\13\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 10+\frac{2}{5}\times 13\\\frac{4}{11}\times 10-\frac{3}{11}\times 13\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5}\\\frac{1}{11}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{1}{5},b=\frac{1}{11}$$

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.

$$30a+44b=10,40a+55b=13$$

To make 30a 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 30.

$$40\times 30a+40\times 44b=40\times 10,30\times 40a+30\times 55b=30\times 13$$

Simplify.

$$1200a+1760b=400,1200a+1650b=390$$

Subtract 1200a+1650b=390 from 1200a+1760b=400 by subtracting like terms on each side of the equal sign.

$$1200a-1200a+1760b-1650b=400-390$$

Add 1200a to -1200a. Terms 1200a and -1200a cancel out, leaving an equation with only one variable that can be solved.

$$1760b-1650b=400-390$$

Add 1760b to -1650b.

$$110b=400-390$$

Add 400 to -390.

$$110b=10$$

Divide both sides by 110.

$$b=\frac{1}{11}$$

Substitute \frac{1}{11} for b in 40a+55b=13. Because the resulting equation contains only one variable, you can solve for a directly.

$$40a+55\times \frac{1}{11}=13$$

Multiply 55 times \frac{1}{11}.

$$40a+5=13$$

Subtract 5 from both sides of the equation.

$$40a=8$$

Divide both sides by 40.

$$a=\frac{1}{5}$$

The system is now solved.

$$a=\frac{1}{5},b=\frac{1}{11}$$