Solve {l}{5a+25b=99}{25a+165b=523}

$$\left. \begin{array} { l } { 5 a + 25 b = 99 } \\ { 25 a + 165 b = 523 } \end{array} \right.$$

Solve for a, b

$$a = \frac{163}{10} = 16\frac{3}{10} = 16.3$$

$$b=\frac{7}{10}=0.7$$

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.

$$5a+25b=99,25a+165b=523$$

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

$$5a+25b=99$$

Subtract 25b from both sides of the equation.

$$5a=-25b+99$$

Divide both sides by 5.

$$a=\frac{1}{5}\left(-25b+99\right)$$

Multiply \frac{1}{5} times -25b+99.

$$a=-5b+\frac{99}{5}$$

Substitute -5b+\frac{99}{5} for a in the other equation, 25a+165b=523.

$$25\left(-5b+\frac{99}{5}\right)+165b=523$$

Multiply 25 times -5b+\frac{99}{5}.

$$-125b+495+165b=523$$

Add -125b to 165b.

$$40b+495=523$$

Subtract 495 from both sides of the equation.

$$40b=28$$

Divide both sides by 40.

$$b=\frac{7}{10}$$

Substitute \frac{7}{10} for b in a=-5b+\frac{99}{5}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-5\times \frac{7}{10}+\frac{99}{5}$$

Multiply -5 times \frac{7}{10}.

$$a=-\frac{7}{2}+\frac{99}{5}$$

Add \frac{99}{5} to -\frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=\frac{163}{10}$$

The system is now solved.

$$a=\frac{163}{10},b=\frac{7}{10}$$

Steps Using Matrices

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

$$5a+25b=99,25a+165b=523$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&25\\25&165\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}99\\523\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&25\\25&165\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&25\\25&165\end{matrix}\right))\left(\begin{matrix}5&25\\25&165\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&25\\25&165\end{matrix}\right))\left(\begin{matrix}99\\523\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&25\\25&165\end{matrix}\right))\left(\begin{matrix}99\\523\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&25\\25&165\end{matrix}\right))\left(\begin{matrix}99\\523\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{165}{5\times 165-25\times 25}&-\frac{25}{5\times 165-25\times 25}\\-\frac{25}{5\times 165-25\times 25}&\frac{5}{5\times 165-25\times 25}\end{matrix}\right)\left(\begin{matrix}99\\523\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{33}{40}&-\frac{1}{8}\\-\frac{1}{8}&\frac{1}{40}\end{matrix}\right)\left(\begin{matrix}99\\523\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{33}{40}\times 99-\frac{1}{8}\times 523\\-\frac{1}{8}\times 99+\frac{1}{40}\times 523\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{163}{10}\\\frac{7}{10}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=\frac{163}{10},b=\frac{7}{10}$$

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.

$$5a+25b=99,25a+165b=523$$

To make 5a and 25a equal, multiply all terms on each side of the first equation by 25 and all terms on each side of the second by 5.

$$25\times 5a+25\times 25b=25\times 99,5\times 25a+5\times 165b=5\times 523$$

Simplify.

$$125a+625b=2475,125a+825b=2615$$

Subtract 125a+825b=2615 from 125a+625b=2475 by subtracting like terms on each side of the equal sign.

$$125a-125a+625b-825b=2475-2615$$

Add 125a to -125a. Terms 125a and -125a cancel out, leaving an equation with only one variable that can be solved.

$$625b-825b=2475-2615$$

Add 625b to -825b.

$$-200b=2475-2615$$

Add 2475 to -2615.

$$-200b=-140$$

Divide both sides by -200.

$$b=\frac{7}{10}$$

Substitute \frac{7}{10} for b in 25a+165b=523. Because the resulting equation contains only one variable, you can solve for a directly.

$$25a+165\times \frac{7}{10}=523$$

Multiply 165 times \frac{7}{10}.

$$25a+\frac{231}{2}=523$$

Subtract \frac{231}{2} from both sides of the equation.

$$25a=\frac{815}{2}$$

Divide both sides by 25.

$$a=\frac{163}{10}$$

The system is now solved.

$$a=\frac{163}{10},b=\frac{7}{10}$$