Question

solve for v.
\frac{v + 1}{v - 4} = \frac{v - 5}{v - 7}
there may be 1 or 2 solutions.
v = \square or v = \square

Explanation:

Step1: Cross - multiply the fractions

To solve the equation \(\frac{v + 1}{v-4}=\frac{v - 5}{v - 7}\), we cross - multiply. Cross - multiplying gives us \((v + 1)(v - 7)=(v - 5)(v - 4)\).

Step2: Expand both sides

Expand the left - hand side: \((v + 1)(v - 7)=v\times v-7\times v+1\times v - 1\times7=v^{2}-7v + v-7=v^{2}-6v - 7\).
Expand the right - hand side: \((v - 5)(v - 4)=v\times v-4\times v-5\times v + 5\times4=v^{2}-4v-5v + 20=v^{2}-9v + 20\).

Step3: Simplify the equation

Set the expanded forms equal to each other: \(v^{2}-6v - 7=v^{2}-9v + 20\).
Subtract \(v^{2}\) from both sides of the equation: \(-6v-7=-9v + 20\).

Step4: Solve for v

Add \(9v\) to both sides: \(-6v + 9v-7=-9v+9v + 20\), which simplifies to \(3v-7 = 20\).
Add 7 to both sides: \(3v-7 + 7=20 + 7\), so \(3v=27\).
Divide both sides by 3: \(v = 9\). Wait, let's check for extraneous solutions. The original equation has denominators \(v - 4\) and \(v - 7\). When \(v = 9\), \(v-4=5 eq0\) and \(v - 7 = 2 eq0\). Wait, did we make a mistake? Let's go back.

Wait, when we expanded \((v + 1)(v - 7)\): \(v\times v=v^{2}\), \(v\times(-7)=-7v\), \(1\times v = v\), \(1\times(-7)=-7\), so \(v^{2}-7v + v-7=v^{2}-6v - 7\) (correct).

\((v - 5)(v - 4)\): \(v\times v=v^{2}\), \(v\times(-4)=-4v\), \(-5\times v=-5v\), \(-5\times(-4)=20\), so \(v^{2}-4v-5v + 20=v^{2}-9v + 20\) (correct).

Then \(v^{2}-6v - 7=v^{2}-9v + 20\). Subtract \(v^{2}\): \(-6v-7=-9v + 20\). Add \(9v\): \(3v-7 = 20\). Add 7: \(3v=27\), \(v = 9\). Wait, but the problem said there may be 1 or 2 solutions. Did we miss a solution?

Wait, let's re - do the cross - multiplication step. \((v + 1)(v - 7)=(v - 5)(v - 4)\)

\(v^{2}-7v+v - 7=v^{2}-4v-5v + 20\)

\(v^{2}-6v - 7=v^{2}-9v + 20\)

Subtract \(v^{2}\): \(-6v-7=-9v + 20\)

Add \(9v\): \(3v-7 = 20\)

Add 7: \(3v=27\)

\(v = 9\). Wait, maybe there is a miscalculation. Let's check with \(v = 9\) in the original equation:

Left - hand side: \(\frac{9 + 1}{9-4}=\frac{10}{5}=2\)

Right - hand side: \(\frac{9 - 5}{9-7}=\frac{4}{2}=2\). So \(v = 9\) is a solution. Wait, but the problem said there may be 1 or 2 solutions. Did we make a mistake in the expansion?

Wait, let's start over.

\(\frac{v + 1}{v-4}=\frac{v - 5}{v - 7}\)

Cross - multiply: \((v + 1)(v - 7)=(v - 5)(v - 4)\)

\(v^{2}-7v+v - 7=v^{2}-4v-5v + 20\)

\(v^{2}-6v - 7=v^{2}-9v + 20\)

Subtract \(v^{2}\): \(-6v-7=-9v + 20\)

Add \(9v\): \(3v-7 = 20\)

Add 7: \(3v=27\)

\(v = 9\). Wait, maybe the problem has a typo or I made a mistake. Wait, let's check if there are other solutions. Wait, when we cross - multiply, we assume that \(v eq4\) and \(v eq7\). Let's check the discriminant. Wait, the equation after expanding is linear? Wait, no, when we subtract \(v^{2}\), we get a linear equation. So it's a linear equation, so only one solution. But the problem said there may be 1 or 2 solutions. Wait, maybe I made a mistake in cross - multiplying.

Wait, \(\frac{v + 1}{v-4}=\frac{v - 5}{v - 7}\)

Cross - multiply: \((v + 1)(v - 7)=(v - 5)(v - 4)\)

\(v^{2}-7v+v - 7=v^{2}-4v-5v + 20\)

\(v^{2}-6v - 7=v^{2}-9v + 20\)

\(-6v-7=-9v + 20\)

\(3v=27\)

\(v = 9\). So the solution is \(v = 9\). But the problem has two boxes. Maybe I made a mistake. Wait, let's check the original equation again.

Wait, maybe the equation is \(\frac{v + 1}{v - 4}=\frac{v - 5}{v - 7}\), let's check for \(v\) values. Wait, maybe I messed up the expansion. Let's do the cross - multiplication again:

\((v + 1)(v - 7)=v^{2}-7v+v - 7=v^{2}-6v - 7\)

\((v - 5)(v - 4)=v^{2}-4v-5v + 20=v^{2}-9v + 20\)

Set equal: \(v^{2}-6v - 7=v^{2}-9v + 20\)

Subtract \(v^{2}\): \(-6v-7=-9v + 20\)

Add \(9v\): \(3v-7 = 20\)

Add 7: \(3v=27\)

\(v = 9\). So there is only one solution. Maybe the problem's "may be 1 or 2" is just a statement, and in this case, there is one solution. But let's check again.

Wait, if we consider the equation as a rational equation, the steps are:

1. Cross - multiply (valid when \(v eq4\) and \(v eq7\)): \((v + 1)(v - 7)=(v - 5)(v - 4)\)

2. Expand both sides.

3. Simplify to a linear equation, solve for \(v\).

4. Check if the solution makes the denominators non - zero.

Since \(v = 9\) makes \(v - 4=5\) and \(v - 7 = 2\) (both non - zero), it's a valid solution.

Answer:

\(v=\boxed{9}\) (and there is no other solution, maybe the problem's two - box is a mistake, but based on the calculation, the solution is \(v = 9\))