Question

question 40 of 42
what values of x make the two expressions below equal?
\\(\frac{(x+3)(x+8)}{5(x+8)} = \frac{x+3}{5}\\)
a. all real numbers except -8 and -3
b. all real numbers except -3
c. all real numbers except -8
d. all real numbers

Explanation:

Step1: Analyze the denominator

For a fraction, the denominator cannot be zero. In the left - hand side expression \(\frac{(x + 3)(x+8)}{5(x + 8)}\), the denominator is \(5(x + 8)\). We set the denominator equal to zero and solve for \(x\): \(5(x + 8)=0\), which gives \(x=-8\). So \(x\) cannot be \(-8\) because it would make the original left - hand side fraction undefined.

Step2: Simplify the expression

We can simplify \(\frac{(x + 3)(x + 8)}{5(x + 8)}\) by canceling out the common factor \((x + 8)\) (when \(x
eq - 8\), because when \(x=-8\), we can't cancel as the expression is undefined at \(x = - 8\)). After canceling \((x + 8)\), we get \(\frac{x + 3}{5}\), which is the same as the right - hand side expression. There is no restriction from the simplified expression on \(x=-3\) because when \(x=-3\), both the original left - hand side (after ensuring \(x
eq - 8\)) and the right - hand side are defined: for the left - hand side, when \(x=-3\) and \(x
eq - 8\), \(\frac{(-3 + 3)(-3+8)}{5(-3 + 8)}=\frac{0\times5}{5\times5}=0\), and the right - hand side is \(\frac{-3 + 3}{5}=0\).

Answer:

C. All real numbers except -8