Question

solve for v.
\\(\frac{v + 6}{v + 2} = \frac{v - 3}{v + 5}\\)
there may be 1 or 2 solutions.
\\(v = \square\\) or \\(v = \square\\)

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{v + 6}{v + 2}=\frac{v - 3}{v + 5}\), we use the cross - multiplication property of fractions. If \(\frac{a}{b}=\frac{c}{d}\) (where \(b eq0\) and \(d eq0\)), then \(a\times d=c\times b\). So we get \((v + 6)(v + 5)=(v - 3)(v + 2)\).

Step2: Expand both sides

Expand the left - hand side: \((v + 6)(v + 5)=v\times v+v\times5 + 6\times v+6\times5=v^{2}+5v + 6v+30=v^{2}+11v + 30\).

Expand the right - hand side: \((v - 3)(v + 2)=v\times v+v\times2-3\times v - 3\times2=v^{2}+2v-3v - 6=v^{2}-v - 6\).

Step3: Simplify the equation

Set the expanded forms equal to each other: \(v^{2}+11v + 30=v^{2}-v - 6\).

Subtract \(v^{2}\) from both sides of the equation: \(v^{2}-v^{2}+11v + 30=v^{2}-v^{2}-v - 6\), which simplifies to \(11v+30=-v - 6\).

Step4: Solve for v

Add \(v\) to both sides: \(11v + v+30=-v + v-6\), so \(12v+30=-6\).

Subtract 30 from both sides: \(12v+30 - 30=-6 - 30\), so \(12v=-36\).

Divide both sides by 12: \(v=\frac{-36}{12}=-3\).

We need to check for extraneous solutions. The original equation has denominators \(v + 2\) and \(v + 5\). When \(v=-3\), \(v + 2=-3 + 2=-1 eq0\) and \(v + 5=-3 + 5 = 2 eq0\).

Wait, let's re - check the expansion and solving process.

Starting again from \((v + 6)(v + 5)=(v - 3)(v + 2)\)

Left side: \(v^{2}+5v+6v + 30=v^{2}+11v + 30\)

Right side: \(v^{2}+2v-3v - 6=v^{2}-v - 6\)

Subtract right side from left side: \(v^{2}+11v + 30-(v^{2}-v - 6)=0\)

\(v^{2}+11v + 30 - v^{2}+v + 6 = 0\)

\(12v+36 = 0\)

\(12v=-36\)

\(v=-3\)

Wait, I made a mistake in the sign when subtracting. Let's do it correctly:

\(v^{2}+11v + 30=v^{2}-v - 6\)

Subtract \(v^{2}\) from both sides: \(11v+30=-v - 6\)

Add \(v\) to both sides: \(12v+30=-6\)

Subtract 30: \(12v=-36\)

\(v = - 3\)

Wait, but let's check the original equation:

Left side: \(\frac{-3 + 6}{-3+2}=\frac{3}{-1}=-3\)

Right side: \(\frac{-3 - 3}{-3 + 5}=\frac{-6}{2}=-3\)

So it works. But wait, maybe I missed a solution? Let's go back to the cross - multiplication step.

The equation is \(\frac{v + 6}{v + 2}=\frac{v - 3}{v + 5}\)

Cross - multiply: \((v + 6)(v + 5)-(v - 3)(v + 2)=0\)

Expand \((v + 6)(v + 5)=v^{2}+11v + 30\)

Expand \((v - 3)(v + 2)=v^{2}-v - 6\)

Subtract: \(v^{2}+11v + 30-(v^{2}-v - 6)=v^{2}+11v + 30 - v^{2}+v + 6=12v + 36=0\)

\(12v=-36\)

\(v=-3\)

So there is only one solution? But the problem said there may be 1 or 2 solutions. Wait, maybe I made a mistake in the cross - multiplication. Let's check again.

\(\frac{v + 6}{v + 2}=\frac{v - 3}{v + 5}\)

Cross - multiply: \((v + 6)(v + 5)=(v - 3)(v + 2)\)

\(v^{2}+5v+6v + 30=v^{2}+2v-3v - 6\)

\(v^{2}+11v + 30=v^{2}-v - 6\)

Subtract \(v^{2}\) from both sides: \(11v+30=-v - 6\)

Add \(v\) to both sides: \(12v+30=-6\)

Subtract 30: \(12v=-36\)

\(v=-3\)

Yes, that's correct. So the solution is \(v=-3\). Wait, but maybe the problem was written incorrectly? Or maybe I misread it. Let me check the original equation again: \(\frac{v + 6}{v + 2}=\frac{v - 3}{v + 5}\). Yes. So cross - multiplying gives a linear equation, so only one solution.

Wait, maybe I made a mistake in the expansion. Let's do \((v + 6)(v + 5)=v^{2}+5v+6v + 30=v^{2}+11v + 30\)

\((v - 3)(v + 2)=v^{2}+2v-3v - 6=v^{2}-v - 6\)

So \(v^{2}+11v + 30=v^{2}-v - 6\)

Subtract \(v^{2}\) from both sides: \(11v+30=-v - 6\)

\(12v=-36\)

\(v=-3\)

And when we check, it works. So the solution is \(v=-3\). But the problem said "there may be 1 or 2 solutions". Maybe in my first calculation, I had a sign error, but after re - checking, it's correct.

Wait, let's consider if the equation was \(\frac{v + 6}{v - 2}=\frac{v - 3}{v + 5}\), but no, the original is \(v + 2\) and \(v + 5\). So the only solution is \(v=-3\).

Answer:

\(v=-3\) (Note: After re - checking the calculation, there is only one solution. Maybe the problem's hint about 1 or 2 solutions is just a general statement. The correct solution is \(v = - 3\))