Solve {l}{10A-8C=+132}{317-5C=-180}

$$\left. \begin{array} { l } { 10 A - 8 C = + 132 } \\ { 317 - 5 C = - 180 } \end{array} \right.$$

Solve for A, C

$$A = \frac{2318}{25} = 92\frac{18}{25} = 92.72$$

$$C = \frac{497}{5} = 99\frac{2}{5} = 99.4$$

Solution Steps

Consider the second equation. Subtract 317 from both sides.

$$-5C=-180-317$$

Subtract 317 from -180 to get -497.

$$-5C=-497$$

Divide both sides by -5.

$$C=\frac{-497}{-5}$$

Fraction \frac{-497}{-5} can be simplified to \frac{497}{5} by removing the negative sign from both the numerator and the denominator.

$$C=\frac{497}{5}$$

Consider the first equation. Insert the known values of variables into the equation.

$$10A-8\times \frac{497}{5}=132$$

Multiply -8 and \frac{497}{5} to get -\frac{3976}{5}.

$$10A-\frac{3976}{5}=132$$

Add \frac{3976}{5} to both sides.

$$10A=132+\frac{3976}{5}$$

Add 132 and \frac{3976}{5} to get \frac{4636}{5}.

$$10A=\frac{4636}{5}$$

Divide both sides by 10.

$$A=\frac{\frac{4636}{5}}{10}$$

Express \frac{\frac{4636}{5}}{10} as a single fraction.

$$A=\frac{4636}{5\times 10}$$

Multiply 5 and 10 to get 50.

$$A=\frac{4636}{50}$$

Reduce the fraction \frac{4636}{50} to lowest terms by extracting and canceling out 2.

$$A=\frac{2318}{25}$$

The system is now solved.

$$A=\frac{2318}{25} C=\frac{497}{5}$$