Question

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Explanation:

Step1: Simplify left side

Simplify \(\frac{v + 2}{v + 6}-1\). First, write \(1\) as \(\frac{v + 6}{v + 6}\), so:
\[
\frac{v + 2}{v + 6}-\frac{v + 6}{v + 6}=\frac{(v + 2)-(v + 6)}{v + 6}=\frac{v + 2 - v - 6}{v + 6}=\frac{-4}{v + 6}
\]
Now the equation becomes \(\frac{-4}{v + 6}=\frac{v - 3}{v + 7}\).

Step2: Cross - multiply

Cross - multiply to get rid of the fractions: \(-4(v + 7)=(v - 3)(v + 6)\).

Step3: Expand both sides

Expand the left side: \(-4v-28\).
Expand the right side: \(v^{2}+6v-3v - 18=v^{2}+3v - 18\).
So the equation is \(-4v-28 = v^{2}+3v - 18\).

Step4: Rearrange into standard quadratic form

Bring all terms to one side: \(v^{2}+3v - 18+4v + 28 = 0\), which simplifies to \(v^{2}+7v + 10 = 0\).

Step5: Factor the quadratic

Factor \(v^{2}+7v + 10\). We need two numbers that multiply to \(10\) and add to \(7\). Those numbers are \(2\) and \(5\). So, \(v^{2}+7v + 10=(v + 2)(v + 5)=0\).

Step6: Solve for v

Set each factor equal to zero:

• If \(v + 2 = 0\), then \(v=-2\).
• If \(v + 5 = 0\), then \(v=-5\).

Step7: Check for extraneous solutions

We need to check if these solutions make the original denominators zero.
For \(v=-2\):

• Denominator \(v + 6=-2 + 6 = 4

eq0\).

• Denominator \(v + 7=-2 + 7 = 5

eq0\).

For \(v=-5\):

• Denominator \(v + 6=-5 + 6 = 1

eq0\).

• Denominator \(v + 7=-5 + 7 = 2

eq0\).

Both solutions are valid.

Answer:

\(-2,-5\)