Question

solve for p. \\(\frac{p - 7}{p + 5} = \frac{p + 5}{p + 1}\\) there may be 1 or 2 solutions. \\(p = \square\\) or \\(p = \square\\)

Explanation:

Step1: Cross - multiply the fractions

Given the equation \(\frac{p - 7}{p + 5}=\frac{p + 5}{p + 1}\), cross - multiplying (which is valid when \(p eq - 5\) and \(p eq - 1\)) gives \((p - 7)(p + 1)=(p + 5)(p + 5)\).

Step2: Expand both sides

Expand the left - hand side: \((p - 7)(p + 1)=p\times p+p\times1-7\times p - 7\times1=p^{2}+p - 7p - 7=p^{2}-6p - 7\).
Expand the right - hand side: \((p + 5)(p + 5)=p^{2}+5p+5p + 25=p^{2}+10p + 25\).
So the equation becomes \(p^{2}-6p - 7=p^{2}+10p + 25\).

Step3: Simplify the equation

Subtract \(p^{2}\) from both sides of the equation:
\(p^{2}-6p - 7-p^{2}=p^{2}+10p + 25-p^{2}\)
\(-6p - 7 = 10p+25\).

Step4: Solve for \(p\)

Add \(6p\) to both sides:
\(-6p - 7+6p=10p + 25+6p\)
\(-7 = 16p+25\).
Subtract 25 from both sides:
\(-7-25=16p+25 - 25\)
\(-32 = 16p\).
Divide both sides by 16: \(p=\frac{-32}{16}=-2\).
We also need to check the excluded values. When \(p=-5\), the denominators \(p + 5 = 0\) in the original equation, and when \(p=-1\), the denominator \(p + 1 = 0\) in the original equation. The solution \(p=-2\) does not make any denominator zero. Wait, but let's re - check our cross - multiplication step. Wait, maybe we made a mistake. Wait, let's start over.

Wait, the original equation is \(\frac{p - 7}{p + 5}=\frac{p + 5}{p + 1}\). Cross - multiply: \((p - 7)(p + 1)=(p + 5)^{2}\).

Expanding left side: \(p^{2}+p-7p - 7=p^{2}-6p - 7\).

Expanding right side: \(p^{2}+10p + 25\).

So \(p^{2}-6p - 7=p^{2}+10p + 25\).

Subtract \(p^{2}\) from both sides: \(-6p-7 = 10p + 25\).

Bring all \(p\) terms to one side: \(-6p-10p=25 + 7\).

\(-16p=32\).

\(p=-2\).

Wait, but the problem says there may be 1 or 2 solutions. Wait, maybe we made a mistake in the cross - multiplication. Wait, no, the equation is a proportion. But let's check if we can have another solution. Wait, maybe the equation is a quadratic, but when we simplified, we got a linear equation. Let's check the discriminant. Wait, when we have \((p - 7)(p + 1)-(p + 5)^{2}=0\).

\(p^{2}-6p - 7-(p^{2}+10p + 25)=0\).

\(p^{2}-6p - 7 - p^{2}-10p - 25=0\).

\(-16p-32 = 0\).

\(-16p=32\).

\(p=-2\).

Wait, but maybe the original equation is \(\frac{p - 7}{p + 5}=\frac{p + 5}{p - 1}\) (a typo?), but according to the given problem, it's \(p + 1\). So with the given problem, the solution is \(p=-2\). But the problem says "there may be 1 or 2 solutions". Wait, maybe we made a mistake. Wait, let's check the original equation again.

Wait, if we consider the equation \(\frac{p - 7}{p + 5}=\frac{p + 5}{p + 1}\), cross - multiply: \((p - 7)(p + 1)=(p + 5)^{2}\).

\(p^{2}-6p - 7=p^{2}+10p + 25\).

\(-16p=32\), so \(p=-2\).

But maybe the problem was written incorrectly. Alternatively, maybe we misread the equation. Wait, if the equation was \(\frac{p - 7}{p - 5}=\frac{p + 5}{p + 1}\), then the solution would be different. But according to the given problem, the denominator is \(p + 5\) and \(p + 1\). So the only solution is \(p=-2\). But the problem has two boxes, so maybe there is a mistake in our calculation. Wait, let's check again.

Wait, let's substitute \(p=-2\) into the original equation:

Left - hand side: \(\frac{-2 - 7}{-2 + 5}=\frac{-9}{3}=-3\).

Right - hand side: \(\frac{-2 + 5}{-2 + 1}=\frac{3}{-1}=-3\). So it works.

Wait, maybe the problem was supposed to be \(\frac{p - 7}{p - 5}=\frac{p + 5}{p + 1}\). Let's try that. Cross - multiply: \((p - 7)(p + 1)=(p - 5)(p + 5)\).

Left side: \(p^{2}+p-7p - 7=p^{2}-6p - 7\).

Right side: \(p^{2}-25\).

So \(p^{2}-6p - 7=p^{2}-25\).

Subtract \(p^{2}\): \(-6p - 7=-25\).

\(-6p=-18\), \(p = 3\). But that's not the case here.

Alternatively, maybe the original equation is \(\frac{p + 7}{p + 5}=\frac{p + 5}{p + 1}\). Let's try that. Cross - multiply: \((p + 7)(p + 1)=(p + 5)^{2}\).

Left side: \(p^{2}+p+7p + 7=p^{2}+8p + 7\).

Right side: \(p^{2}+10p + 25\).

\(p^{2}+8p + 7=p^{2}+10p + 25\).

\(8p+7 = 10p+25\).

\(-2p=18\), \(p=-9\).

But according to the given problem, it's \(p - 7\). So with the given problem, the solution is \(p=-2\). But the problem has two boxes, so maybe there is a mistake. Wait, maybe we made a mistake in the sign when expanding. Let's re - expand \((p - 7)(p + 1)\):

\(p\times p=p^{2}\), \(p\times1 = p\), \(-7\times p=-7p\), \(-7\times1=-7\). So \(p^{2}+p-7p - 7=p^{2}-6p - 7\). Correct.

\((p + 5)(p + 5)=p^{2}+10p + 25\). Correct.

So \(p^{2}-6p - 7=p^{2}+10p + 25\). Subtract \(p^{2}\): \(-6p-7 = 10p + 25\). Then \(-16p=32\), \(p=-2\).

So the only solution is \(p=-2\). But the problem says "there may be 1 or 2 solutions" and has two boxes. Maybe there is a typo in the problem. But based on the given equation, the solution is \(p=-2\).

Wait, maybe the original equation is \(\frac{p - 7}{p + 5}=\frac{p - 5}{p + 1}\). Let's try that. Cross - multiply: \((p - 7)(p + 1)=(p - 5)(p + 5)\).

Left side: \(p^{2}+p-7p - 7=p^{2}-6p - 7\).

Right side: \(p^{2}-25\).

\(p^{2}-6p - 7=p^{2}-25\).

\(-6p=-18\), \(p = 3\).

But this is not the given equation. So, based on the given equation, the solution is \(p=-2\).

Answer:

\(p=-2\) (Since the problem has two boxes but the equation as given has only one solution, maybe there is a mistake in the problem statement. But based on the calculation, the solution is \(p = - 2\))