Question

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

Explanation:

Step1: Cross - multiply to eliminate fractions

To solve the equation \(\frac{5}{v - 6}=\frac{2}{v}\), we cross - multiply. Cross - multiplying gives us \(5\times v=2\times(v - 6)\). This is based on the property of proportions: if \(\frac{a}{b}=\frac{c}{d}\), then \(a\times d = b\times c\) (where \(b eq0\) and \(d eq0\)). So we have \(5v = 2(v - 6)\).

Step2: Expand the right - hand side

Expand \(2(v - 6)\) using the distributive property \(a(b + c)=ab+ac\). Here, \(a = 2\), \(b=v\) and \(c=-6\), so \(2(v - 6)=2v-12\). The equation becomes \(5v=2v - 12\).

Step3: Subtract \(2v\) from both sides

Subtract \(2v\) from both sides of the equation \(5v=2v - 12\) to get \(5v-2v=2v - 12-2v\). Simplifying the left - hand side: \(5v - 2v = 3v\), and the right - hand side: \(2v-2v-12=-12\). So the equation is \(3v=-12\).

Step4: Solve for \(v\)

Divide both sides of the equation \(3v=-12\) by 3. \(\frac{3v}{3}=\frac{-12}{3}\), which gives \(v=- 4\). Wait, we also need to check for extraneous solutions because we had denominators \(v-6\) and \(v\). The original equation is undefined when \(v = 0\) or \(v = 6\). Let's check our solution \(v=-4\):
For the left - hand side: \(\frac{5}{-4 - 6}=\frac{5}{-10}=-\frac{1}{2}\)
For the right - hand side: \(\frac{2}{-4}=-\frac{1}{2}\)
Wait, maybe I made a mistake in the cross - multiplication step? Wait, no, let's go back. Wait, the original equation is \(\frac{5}{v - 6}=\frac{2}{v}\). Cross - multiplying gives \(5v=2(v - 6)\), which is \(5v=2v-12\), then \(5v - 2v=-12\), \(3v=-12\), \(v = - 4\). But wait, is there another solution? Wait, maybe I made a mistake in the problem - solving process. Wait, let's start over.

Step1 (revised): Cross - multiply

\(\frac{5}{v - 6}=\frac{2}{v}\)
Cross - multiply: \(5v=2(v - 6)\)

Step2 (revised): Expand

\(5v=2v-12\)

Step3 (revised): Subtract \(2v\) from both sides

\(5v-2v=2v-12 - 2v\)
\(3v=-12\)

Step4 (revised): Divide by 3

\(v=-4\)
Wait, but the problem says there may be 1 or 2 solutions. Did I miss something? Wait, no, let's check the domain. The original equation has \(v eq0\) and \(v eq6\). The solution \(v = - 4\) is valid. Wait, maybe the problem was written incorrectly? Or maybe I made a mistake. Wait, let's check the cross - multiplication again. \(\frac{5}{v - 6}=\frac{2}{v}\) implies \(5v=2(v - 6)\), which is correct. So \(5v=2v-12\), \(3v=-12\), \(v=-4\). So there is only one solution? But the problem says "1 or 2 solutions". Maybe there was a typo in the problem, or maybe I misread the equation. Wait, the equation is \(\frac{5}{v - 6}=\frac{2}{v}\)? Or is it \(\frac{5}{v}+6=\frac{2}{v}\)? No, the user wrote \(\frac{5}{v - 6}=\frac{2}{v}\). So according to the solution, \(v=-4\) is the only solution. But the problem has two boxes. Maybe I made a mistake. Wait, let's solve it again:

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

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

\(5v=2v-12\)

\(5v-2v=-12\)

\(3v=-12\)

\(v=-4\)

Yes, that's correct. So the only solution is \(v=-4\). But the problem has two boxes. Maybe the original equation was \(\frac{5}{v - 6}=\frac{2}{v - something}\)? Or maybe a different equation. Wait, maybe the user made a typo. But based on the given equation \(\frac{5}{v - 6}=\frac{2}{v}\), the solution is \(v=-4\).

Wait, maybe I made a mistake in the sign. Let's check:

Left - hand side when \(v=-4\): \(\frac{5}{-4-6}=\frac{5}{-10}=-\frac{1}{2}\)

Right - hand side when \(v=-4\): \(\frac{2}{-4}=-\frac{1}{2}\). So it's correct. So there is only one solution. But the problem says "1 or 2 solutions". Maybe the equation was supposed to be \(\frac{5}{v - 6}=\frac{2}{v + something}\) or a quadratic equation. Wait, maybe the original equation is \(\frac{5}{v - 6}=\frac{2}{v}\) is a linear equation in terms of \(v\) (after cross - multiplying), so it's a linear equation, so only one solution.

Answer:

\(v=-4\) (and there is no other solution, so maybe the problem has a typo, but based on the given equation, the solution is \(v = - 4\))