Solve the equation for $v$:
$-\frac{8}{v+3}=\frac{5}{4 v+12}+1$
To solve the equation \[-\frac{8}{v+3} = \frac{5}{4v+12} + 1,\] we first examine the denominators. The term \(\frac{5}{4v+12}\) can be rewritten by factoring out a 4 in the denominator: \[4v + 12 = 4(v+3).\] Thus, the equation becomes \[-\frac{8}{v+3} = \frac{5}{4(v+3)} + 1.\] We must note that for the expression to be defined, the denominator cannot be zero, which means \(v+3 eq 0\), or \(v eq -3\).
To clear the fractions, we multiply every term by the least common denominator, which is \(4(v+3)\).
Multiplying the left side:
\[4(v+3) \times \left(-\frac{8}{v+3}\right) = -32.\]
Multiplying the right side:
\[4(v+3) \times \left(\frac{5}{4(v+3)} + 1\right) = 4(v+3) \times \frac{5}{4(v+3)} + 4(v+3) \times 1 = 5 + 4(v+3).\]
Now, we equate the results to form a linear equation:
\[-32 = 5 + 4v + 12.\]
Now we solve the linear equation \(-32 = 4v + 17\).
Subtract \(17\) from both sides:
\[-32 - 17 = 4v\]
\[-49 = 4v\]
Divide by 4:
\[v = -\frac{49}{4}.\]
Since \(-\frac{49}{4} = -12.25\), this is not equal to \(-3\), so our solution is valid.
In various engineering applications, such as calculating the equilibrium point in a balanced bridge structure or determining specific electrical resistance values, rational equations often arise. For instance, when designing a circuit where resistances are combined in parallel or series, you might encounter equations where variables are in the denominator. Solving these correctly ensures that the equipment operates within the intended parameters, preventing failure or inefficient energy usage.