What is the solution set of the equation $\frac{6}{m}+2=\frac{m+3}{m-1}$ ?
Note: $m
eq\{0,1\}$
The objective is to solve the rational equation \(\frac{6}{m} + 2 = \frac{m + 3}{m - 1}\) for \(m\), under the constraints \(m eq 0\) and \(m eq 1\). To solve this, we first need to eliminate the denominators by multiplying both sides of the equation by the least common multiple of the denominators, which is \(m(m - 1)\).
Multiplying both sides by \(m(m - 1)\), we obtain:
\[6(m - 1) + 2m(m - 1) = m(m + 3)\]
Expanding the terms on both sides gives:
\[(6m - 6) + (2m^2 - 2m) = m^2 + 3m\]
Combining like terms on the left side:
\[2m^2 + 4m - 6 = m^2 + 3m\]
Now, move all terms to one side of the equation to set it to zero:
\[2m^2 - m^2 + 4m - 3m - 6 = 0\]
\[m^2 + m - 6 = 0\]
The resulting quadratic equation is \(m^2 + m - 6 = 0\). We can solve this by factoring. We look for two numbers that multiply to \(-6\) and add to \(1\). These numbers are \(3\) and \(-2\). Thus, the equation factors as:
\[(m + 3)(m - 2) = 0\]
Setting each factor to zero gives the solutions:
\[m + 3 = 0 \implies m = -3\]
\[m - 2 = 0 \implies m = 2\]
Since neither \(-3\) nor \(2\) violates the initial constraints \(m
eq 0\) and \(m
eq 1\), both are valid solutions. The solution set is \(\{-3, 2\}\).
Rational equations appear frequently in problems involving rates, such as travel time or work completion. For instance, if you are calculating how long it takes for two machines working together to complete a task, you often encounter equations with variables in the denominators. Solving these accurately helps in optimizing production schedules, managing logistics, or even determining the efficiency of resource allocation in a system. By reducing these to simpler polynomials, you can precisely identify the required variables to meet your operational goals.