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

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

Solve for a, b, c, d

$$d=11$$

Solution Steps

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

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

Calculate 2 to the power of 2 and get 4.

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

Add 4 and 3 to get 7.

$$c=2\times 7+3-2\times 3$$

Multiply 2 and 7 to get 14.

$$c=14+3-2\times 3$$

Add 14 and 3 to get 17.

$$c=17-2\times 3$$

Multiply -2 and 3 to get -6.

$$c=17-6$$

Subtract 6 from 17 to get 11.

$$c=11$$

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

$$d=11$$

The system is now solved.

$$a=2 b=3 c=11 d=11$$