Question

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

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{v + 5}{v + 3}=\frac{v + 3}{v - 1}\), we cross - multiply. That is, \((v + 5)(v - 1)=(v + 3)(v + 3)\).

Step2: Expand both sides

Expand the left - hand side using the FOIL method: \((v+5)(v - 1)=v\times v-v\times1 + 5\times v-5\times1=v^{2}-v + 5v-5=v^{2}+4v - 5\).
Expand the right - hand side using the formula \((a + b)^{2}=a^{2}+2ab + b^{2}\) (here \(a = v\) and \(b = 3\)): \((v + 3)(v + 3)=v^{2}+6v + 9\).

Step3: Set up the equation and simplify

Now we have the equation \(v^{2}+4v - 5=v^{2}+6v + 9\).
Subtract \(v^{2}\) from both sides of the equation: \(v^{2}-v^{2}+4v - 5=v^{2}-v^{2}+6v + 9\), which simplifies to \(4v-5 = 6v + 9\).

Step4: Solve for v

Subtract \(4v\) from both sides: \(4v-4v-5=6v - 4v+9\), so \(-5 = 2v+9\).
Subtract 9 from both sides: \(-5-9=2v+9 - 9\), which gives \(-14 = 2v\).
Divide both sides by 2: \(v=\frac{-14}{2}=-7\).

We need to check for extraneous solutions. The original equation has denominators \(v + 3\) and \(v - 1\). When \(v=-7\), \(v + 3=-7 + 3=-4 eq0\) and \(v - 1=-7-1=-8 eq0\).

Wait, let's check our expansion again. Wait, when we cross - multiplied, we have \((v + 5)(v - 1)=(v + 3)^{2}\).

Expanding \((v + 5)(v - 1)=v^{2}-v+5v - 5=v^{2}+4v - 5\)

Expanding \((v + 3)^{2}=v^{2}+6v + 9\)

Then \(v^{2}+4v - 5=v^{2}+6v + 9\)

Subtract \(v^{2}\) from both sides: \(4v-5 = 6v + 9\)

Subtract \(4v\) from both sides: \(-5=2v + 9\)

Subtract 9: \(-14 = 2v\), so \(v=-7\)

Wait, but let's check if there was a miscalculation. Wait, maybe we made a mistake in the cross - multiplication? No, cross - multiplication for \(\frac{a}{b}=\frac{c}{d}\) (where \(b eq0\) and \(d eq0\)) gives \(ad = bc\). So that step is correct.

Wait, but let's substitute \(v=-7\) into the original equation:

Left - hand side: \(\frac{-7 + 5}{-7+3}=\frac{-2}{-4}=\frac{1}{2}\)

Right - hand side: \(\frac{-7 + 3}{-7-1}=\frac{-4}{-8}=\frac{1}{2}\)

So \(v = - 7\) is a valid solution. But the problem said there may be 1 or 2 solutions. Wait, maybe we missed a solution? Wait, let's go back to the equation \((v + 5)(v - 1)=(v + 3)^{2}\)

\(v^{2}+4v-5=v^{2}+6v + 9\)

\(v^{2}-v^{2}+4v-6v=9 + 5\)

\(-2v=14\)

\(v=-7\)

Wait, that's the same as before. So there is only one solution? But the problem has two boxes. Wait, maybe I made a mistake in the cross - multiplication. Wait, the original equation is \(\frac{v + 5}{v + 3}=\frac{v + 3}{v - 1}\). Let's cross - multiply: \((v + 5)(v - 1)=(v + 3)(v + 3)\). Wait, maybe the user made a typo, or maybe I made a mistake. Wait, let's re - do the algebra:

\((v + 5)(v - 1)=(v + 3)^{2}\)

\(v^{2}-v+5v - 5=v^{2}+6v + 9\)

\(v^{2}+4v-5=v^{2}+6v + 9\)

Subtract \(v^{2}\) from both sides: \(4v-5=6v + 9\)

\(4v-6v=9 + 5\)

\(-2v=14\)

\(v=-7\)

So the solution is \(v=-7\). But the problem has two boxes. Wait, maybe there was an error in the problem statement, or maybe I made a mistake. Wait, let's check if the equation is \(\frac{v + 5}{v - 3}=\frac{v + 3}{v - 1}\) (a different denominator). If the denominator was \(v - 3\) instead of \(v + 3\), let's see:

Cross - multiply: \((v + 5)(v - 1)=(v + 3)(v - 3)\)

\(v^{2}-v+5v - 5=v^{2}-9\)

\(v^{2}+4v-5=v^{2}-9\)

Subtract \(v^{2}\): \(4v-5=-9\)

\(4v=-4\)

\(v=-1\). But that's not the case here.

Wait, in our original problem, the denominator is \(v + 3\). So when we solved, we got \(v=-7\). Let's check again:

Left side: \(\frac{-7 + 5}{-7+3}=\frac{-2}{-4}=\frac{1}{2}\)

Right side: \(\frac{-7 + 3}{-7-1}=\frac{-4}{-8}=\frac{1}{2}\). So it works. So maybe the problem has a typo, or maybe I misread it. Wait, the original equation is \(\frac{v + 5}{v + 3}=\frac{v + 3}{v - 1}\). So the solution is \(v=-7\). But the problem has two boxes. Wait, maybe we made a mistake in the sign when expanding. Wait, \((v + 5)(v - 1)=v^{2}-v + 5v-5=v^{2}+4v - 5\), correct. \((v + 3)^{2}=v^{2}+6v + 9\), correct. Then \(v^{2}+4v - 5-v^{2}-6v - 9 = 0\), so \(-2v-14 = 0\), so \(-2v=14\), \(v=-7\). So there is only one solution. Maybe the problem expects us to consider that when we cross - multiply, we might have a quadratic equation, but in this case, the \(v^{2}\) terms cancel out, resulting in a linear equation, so only one solution.

Answer:

\(v=-7\) (and maybe the second box is a mistake, but the solution is \(v = - 7\))