Solve {l}{17=5a+47b}{192=47a+505b}

$$\left. \begin{array} { l } { 17 = 5 a + 47 b } \\ { 192 = 47 a + 505 b } \end{array} \right.$$

Solve for a, b

$$a = -\frac{439}{316} = -1\frac{123}{316} \approx -1.389240506$$

$$b=\frac{161}{316}\approx 0.509493671$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$5a+47b=17$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$47a+505b=192$$

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.

$$5a+47b=17,47a+505b=192$$

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

$$5a+47b=17$$

Subtract 47b from both sides of the equation.

$$5a=-47b+17$$

Divide both sides by 5.

$$a=\frac{1}{5}\left(-47b+17\right)$$

Multiply \frac{1}{5} times -47b+17.

$$a=-\frac{47}{5}b+\frac{17}{5}$$

Substitute \frac{-47b+17}{5} for a in the other equation, 47a+505b=192.

$$47\left(-\frac{47}{5}b+\frac{17}{5}\right)+505b=192$$

Multiply 47 times \frac{-47b+17}{5}.

$$-\frac{2209}{5}b+\frac{799}{5}+505b=192$$

Add -\frac{2209b}{5} to 505b.

$$\frac{316}{5}b+\frac{799}{5}=192$$

Subtract \frac{799}{5} from both sides of the equation.

$$\frac{316}{5}b=\frac{161}{5}$$

Divide both sides of the equation by \frac{316}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$b=\frac{161}{316}$$

Substitute \frac{161}{316} for b in a=-\frac{47}{5}b+\frac{17}{5}. Because the resulting equation contains only one variable, you can solve for a directly.

$$a=-\frac{47}{5}\times \frac{161}{316}+\frac{17}{5}$$

Multiply -\frac{47}{5} times \frac{161}{316} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{7567}{1580}+\frac{17}{5}$$

Add \frac{17}{5} to -\frac{7567}{1580} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$a=-\frac{439}{316}$$

The system is now solved.

$$a=-\frac{439}{316},b=\frac{161}{316}$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$5a+47b=17$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$47a+505b=192$$

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

$$5a+47b=17,47a+505b=192$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&47\\47&505\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}17\\192\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&47\\47&505\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&47\\47&505\end{matrix}\right))\left(\begin{matrix}5&47\\47&505\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&47\\47&505\end{matrix}\right))\left(\begin{matrix}17\\192\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&47\\47&505\end{matrix}\right))\left(\begin{matrix}17\\192\end{matrix}\right)$$

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

$$\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5&47\\47&505\end{matrix}\right))\left(\begin{matrix}17\\192\end{matrix}\right)$$

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.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{505}{5\times 505-47\times 47}&-\frac{47}{5\times 505-47\times 47}\\-\frac{47}{5\times 505-47\times 47}&\frac{5}{5\times 505-47\times 47}\end{matrix}\right)\left(\begin{matrix}17\\192\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{505}{316}&-\frac{47}{316}\\-\frac{47}{316}&\frac{5}{316}\end{matrix}\right)\left(\begin{matrix}17\\192\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{505}{316}\times 17-\frac{47}{316}\times 192\\-\frac{47}{316}\times 17+\frac{5}{316}\times 192\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{439}{316}\\\frac{161}{316}\end{matrix}\right)$$

Extract the matrix elements a and b.

$$a=-\frac{439}{316},b=\frac{161}{316}$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$5a+47b=17$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$47a+505b=192$$

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.

$$5a+47b=17,47a+505b=192$$

To make 5a and 47a equal, multiply all terms on each side of the first equation by 47 and all terms on each side of the second by 5.

$$47\times 5a+47\times 47b=47\times 17,5\times 47a+5\times 505b=5\times 192$$

Simplify.

$$235a+2209b=799,235a+2525b=960$$

Subtract 235a+2525b=960 from 235a+2209b=799 by subtracting like terms on each side of the equal sign.

$$235a-235a+2209b-2525b=799-960$$

Add 235a to -235a. Terms 235a and -235a cancel out, leaving an equation with only one variable that can be solved.

$$2209b-2525b=799-960$$

Add 2209b to -2525b.

$$-316b=799-960$$

Add 799 to -960.

$$-316b=-161$$

Divide both sides by -316.

$$b=\frac{161}{316}$$

Substitute \frac{161}{316} for b in 47a+505b=192. Because the resulting equation contains only one variable, you can solve for a directly.

$$47a+505\times \frac{161}{316}=192$$

Multiply 505 times \frac{161}{316}.

$$47a+\frac{81305}{316}=192$$

Subtract \frac{81305}{316} from both sides of the equation.

$$47a=-\frac{20633}{316}$$

Divide both sides by 47.

$$a=-\frac{439}{316}$$

The system is now solved.

$$a=-\frac{439}{316},b=\frac{161}{316}$$