Solve {l}{a=-2}{b={(a+a^2/a-1)}*{(a+1/2a^2-a)}}{text{Solvefor}ctext{where}}{c=b}

$$\left. \begin{array} { l } { a = -2 }\\ { b = {(a + \frac{a ^ {2}}{a - 1})} \cdot {(\frac{a + 1}{2 a ^ {2} - a})} }\\ { \text{Solve for } c \text{ where} } \\ { c = b } \end{array} \right.$$

Solve for a, b, c

$$c=\frac{1}{3}\approx 0.333333333$$

Solution Steps

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

$$b=\left(-2+\frac{\left(-2\right)^{2}}{-2-1}\right)\times \frac{-2+1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Calculate -2 to the power of 2 and get 4.

$$b=\left(-2+\frac{4}{-2-1}\right)\times \frac{-2+1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Subtract 1 from -2 to get -3.

$$b=\left(-2+\frac{4}{-3}\right)\times \frac{-2+1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Fraction \frac{4}{-3} can be rewritten as -\frac{4}{3} by extracting the negative sign.

$$b=\left(-2-\frac{4}{3}\right)\times \frac{-2+1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Subtract \frac{4}{3} from -2 to get -\frac{10}{3}.

$$b=-\frac{10}{3}\times \frac{-2+1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Add -2 and 1 to get -1.

$$b=-\frac{10}{3}\times \frac{-1}{2\left(-2\right)^{2}-\left(-2\right)}$$

Calculate -2 to the power of 2 and get 4.

$$b=-\frac{10}{3}\times \frac{-1}{2\times 4-\left(-2\right)}$$

Multiply 2 and 4 to get 8.

$$b=-\frac{10}{3}\times \frac{-1}{8-\left(-2\right)}$$

Multiply -1 and -2 to get 2.

$$b=-\frac{10}{3}\times \frac{-1}{8+2}$$

Add 8 and 2 to get 10.

$$b=-\frac{10}{3}\times \frac{-1}{10}$$

Fraction \frac{-1}{10} can be rewritten as -\frac{1}{10} by extracting the negative sign.

$$b=-\frac{10}{3}\left(-\frac{1}{10}\right)$$

Multiply -\frac{10}{3} and -\frac{1}{10} to get \frac{1}{3}.

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

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

$$c=\frac{1}{3}$$

The system is now solved.

$$a=-2 b=\frac{1}{3} c=\frac{1}{3}$$