Question

question 4 of 5
select the correct answer.
which value of p makes this equation true?
\\(\frac{p + 1}{3} = \frac{p + 1}{4}\\)
\\(\bigcirc\\) \\(p = -1\\)
\\(\bigcirc\\) \\(p = 0\\)
\\(\bigcirc\\) \\(p = 1\\)
\\(\bigcirc\\) the equation has no solution.

Explanation:

Step1: Analyze the equation

We have the equation $\frac{p + 1}{3}=\frac{p + 1}{4}$. Let's consider two cases: when $p+1 = 0$ and when $p + 1
eq0$.

Step2: Case 1: $p + 1=0$

If $p+1 = 0$, then $p=- 1$. Substitute $p = - 1$ into the equation:
Left - hand side: $\frac{-1 + 1}{3}=\frac{0}{3}=0$
Right - hand side: $\frac{-1+1}{4}=\frac{0}{4}=0$
So when $p=-1$, both sides of the equation are equal.

Step3: Case 2: $p + 1

eq0$
If $p + 1
eq0$, we can divide both sides of the equation $\frac{p + 1}{3}=\frac{p + 1}{4}$ by $(p + 1)$ (since $p+1
eq0$, division is allowed). We get $\frac{1}{3}=\frac{1}{4}$, which is a false statement. So there are no solutions in this case except when $p + 1=0$ (i.e., $p=-1$).

Answer:

A. $p=-1$