Solve {l}{b+3=1}{a-3=b^2-2b-3}

$$\left\{ \begin{array} { l } { b + 3 = 1 } \\ { a - 3 = b ^ { 2 } - 2 b - 3 } \end{array} \right.$$

Solve for b, a

$$b=-2$$

$$a=8$$

Solution Steps

Consider the first equation. Subtract 3 from both sides.

$$b=1-3$$

Subtract 3 from 1 to get -2.

$$b=-2$$

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

$$a-3=\left(-2\right)^{2}-2\left(-2\right)-3$$

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

$$a-3=4-2\left(-2\right)-3$$

Multiply -2 and -2 to get 4.

$$a-3=4+4-3$$

Add 4 and 4 to get 8.

$$a-3=8-3$$

Subtract 3 from 8 to get 5.

$$a-3=5$$

Add 3 to both sides.

$$a=5+3$$

Add 5 and 3 to get 8.

$$a=8$$

The system is now solved.

$$b=-2 a=8$$