To solve the equation \(\frac{v+6}{v+1}=\frac{v+7}{v+5}+1\), we first need to eliminate the fractions by finding a common denominator. Then, we can simplify the equation and solve for \(v\) by isolating it on one side of the equation. Finally, we will check for any extraneous solutions that might arise from the original equation's restrictions.
We start with the equation:
\[
\frac{v+6}{v+1} = \frac{v+7}{v+5} + 1
\]
To eliminate the fractions, we find a common denominator and rewrite the equation:
\[
\frac{v+6}{v+1} = \frac{v+7 + (v+5)}{v+5}
\]
Simplifying the right side:
\[
\frac{v+6}{v+1} = \frac{2v+12}{v+5}
\]
Cross-multiply to eliminate the denominators:
\[
(v+6)(v+5) = (v+1)(2v+12)
\]
Expand both sides:
\[
v^2 + 5v + 6v + 30 = 2v^2 + 12v + v + 12
\]
Simplify:
\[
v^2 + 11v + 30 = 2v^2 + 13v + 12
\]
Rearrange the terms to form a quadratic equation:
\[
0 = 2v^2 + 13v + 12 - v^2 - 11v - 30
\]
Simplify:
\[
0 = v^2 + 2v - 18
\]
Factor the quadratic equation:
\[
(v + 6)(v - 3) = 0
\]
Set each factor to zero and solve for \(v\):
\[
v + 6 = 0 \quad \Rightarrow \quad v = -6
\]
\[
v - 3 = 0 \quad \Rightarrow \quad v = 3
\]
Check the solutions in the original equation to ensure they do not make any denominator zero:
- For \(v = -6\), the denominators are \(-5\) and \(-1\), which are valid.
- For \(v = 3\), the denominators are \(4\) and \(8\), which are valid.
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