Question

1. solve the equation. show all your work. *

$x + 2 + x^2 - (x - 2 + 1) = x^2 + 3 + 3x - (2x + 2)$
a. $x = 1$
b. $x = 2$
c. $x = -2$
d. no solution

Explanation:

Step1: Simplify both sides

First, simplify the left - hand side (LHS):
\[

$$\begin{align*} x + 2+x^{2}-(x - 2 + 1)&=x^{2}+x + 2-(x - 1)\\ &=x^{2}+x + 2 - x+1\\ &=x^{2}+3 \end{align*}$$

\]
Then, simplify the right - hand side (RHS):
\[

$$\begin{align*} x^{2}+3 + 3x-(2x + 2)&=x^{2}+3+3x - 2x-2\\ &=x^{2}+x + 1 \end{align*}$$

\]

Step2: Set up the simplified equation

Now we have the equation \(x^{2}+3=x^{2}+x + 1\)
Subtract \(x^{2}\) from both sides of the equation:
\(x^{2}+3-x^{2}=x^{2}+x + 1-x^{2}\)
Which simplifies to \(3=x + 1\)

Step3: Solve for x

Subtract 1 from both sides:
\(3-1=x+1 - 1\)
\(2=x\)
Wait, but let's check our simplification again. Wait, when we simplified the LHS:
Original LHS: \(x + 2+x^{2}-(x - 2 + 1)=x + 2+x^{2}-(x - 1)=x^{2}+x + 2 - x + 1=x^{2}+3\) (correct)
Original RHS: \(x^{2}+3 + 3x-(2x + 2)=x^{2}+3x + 3-2x - 2=x^{2}+x+1\) (correct)
Then \(x^{2}+3=x^{2}+x + 1\)
Subtract \(x^{2}\) from both sides: \(3=x + 1\)
Subtract 1 from both sides: \(x=2\)

Wait, let's verify by plugging \(x = 2\) into the original equation:
LHS: \(2+2 + 2^{2}-(2 - 2+1)=4 + 4-(1)=4 + 4-1 = 7\)
RHS: \(2^{2}+3+3\times2-(2\times2 + 2)=4 + 3+6-(4 + 2)=13 - 6 = 7\)
LHS = RHS when \(x = 2\)

Answer:

B. \(x = 2\)