Solve {l}{a+4=3}{text{Solvefor}btext{where}}{b=a^3+2a^2}

$$\left. \begin{array} { l } { a + 4 = 3 }\\ { \text{Solve for } b \text{ where} } \\ { b = a ^ {3} + 2 a ^ {2} } \end{array} \right.$$

Solve for a, b

$$b=1$$

Solution Steps

Consider the first equation. Subtract 4 from both sides.

$$a=3-4$$

Subtract 4 from 3 to get -1.

$$a=-1$$

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

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

Calculate -1 to the power of 3 and get -1.

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

Calculate -1 to the power of 2 and get 1.

$$b=-1+2\times 1$$

Multiply 2 and 1 to get 2.

$$b=-1+2$$

Add -1 and 2 to get 1.

$$b=1$$

The system is now solved.

$$a=-1 b=1$$