Solve {l}{a=2}{b=-1}{c={(a+b)}{(a+b)}+{(a-2b)}{(a+3b)}}{text{Solvefor}dtext{where}}{d=c}

$$\left. \begin{array} { l } { a = 2 }\\ { b = -1 }\\ { c = {(a + b)} {(a + b)} + {(a - 2 b)} {(a + 3 b)} }\\ { \text{Solve for } d \text{ where} } \\ { d = c } \end{array} \right.$$

Solve for a, b, c, d

$$d=-3$$

Solution Steps

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

$$c=\left(2-1\right)\left(2-1\right)+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Multiply 2-1 and 2-1 to get \left(2-1\right)^{2}.

$$c=\left(2-1\right)^{2}+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Subtract 1 from 2 to get 1.

$$c=1^{2}+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Calculate 1 to the power of 2 and get 1.

$$c=1+\left(2-2\left(-1\right)\right)\left(2+3\left(-1\right)\right)$$

Multiply -2 and -1 to get 2.

$$c=1+\left(2+2\right)\left(2+3\left(-1\right)\right)$$

Add 2 and 2 to get 4.

$$c=1+4\left(2+3\left(-1\right)\right)$$

Multiply 3 and -1 to get -3.

$$c=1+4\left(2-3\right)$$

Subtract 3 from 2 to get -1.

$$c=1+4\left(-1\right)$$

Multiply 4 and -1 to get -4.

$$c=1-4$$

Subtract 4 from 1 to get -3.

$$c=-3$$

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

$$d=-3$$

The system is now solved.

$$a=2 b=-1 c=-3 d=-3$$