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Magazine Chapter 6: Problem 78
Solve each equation, and check the solutions. $$ \frac{m}{m^{2}+m-2}=\frac{m}{m^{2}+3 m+2}-\frac{m}{m^{2}-1} $$
Short Answer
Expert verified
The solutions are m = 0 and m = -3.
Step by step solution
01
Factor the denominators
First, factor the denominators of each fraction.For the denominator of the first term, notice that:divide denominator \( m^{2}+m-2 = (m+2)(m-1)\) For the denominator of the second term: \(m^{2}+3m+2 = (m+1)(m+2)\)Finally, for the denominator of the third term: \(m^{2}-1 = (m-1)(m+1)\)After factoring, the original equation can be rewritten as: \[\frac{m}{(m+2)(m-1)} = \frac{m}{(m+1)(m+2)} - \frac{m}{(m-1)(m+1)}\]
02
Find a common denominator
To solve the equation, find a common denominator for all three fractions. The common denominator will be \((m+2)(m-1)(m+1))\).Then rewrite each fraction with the common denominator: \[\frac{m}{(m+2)(m-1)} = \frac{m(m+1)}{(m+2)(m-1)(m+1)}\] \[\frac{m}{(m+1)(m+2)} = \frac{m(m-1)}{(m+2)(m-1)(m+1)} \] \[\frac{m}{(m-1)(m+1)} = \frac{m(m+2)}{(m+2)(m-1)(m+1)} \]
03
Simplify the equation
Now substitute back into the original equation: \[ \frac{m(m+1)}{(m+2)(m-1)(m+1)} = \frac{m(m-1)}{(m+2)(m-1)(m+1)} - \frac{m(m+2)}{(m+2)(m-1)(m+1)} \]Notice that all fractions now have the same denominator, so the numerators can be combined:\[m(m+1) = m(m-1) - m(m+2)\]
04
Expand and combine like terms
Expand the numerators:\[ m^{2} + m = m^{2} - m - m^{2} - 2m \] Combine like terms:\[ m^{2} + m = -2m \]then combine like terms to isolate the variable m on one side
05
Solve for m
Combine the like terms on the right-hand side:\[ m^{2}+ 3m = 0 \]Factor m out from the equation \[ m(m + 3) = 0 \]Setting each factor to zero gives us the solutions: \[m = 0 \] and \[ m = -3 \]
06
Check the solutions
Substitute the solutions back into the original equation to verify.For \(m = 0\): The denominators become non-zero and equation balancesFor \(m = -3\): The denominators become non-zero and equation balances.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
factoring polynomials
Factoring polynomials is a crucial first step when solving rational equations. It enables us to break down complex expressions into simpler components. For example, consider the polynomial in the first denominator, \(m^2 + m - 2\). To factor this, we look for two numbers that multiply to \(-2\) and add up to \(1\). These numbers are \(2\) and \(-1\), so we can factor the polynomial as \((m + 2)(m - 1)\). Factoring the other polynomials follows a similar process:
- \(m^2 + 3m + 2 = (m + 1)(m + 2)\)
- \(m^2 - 1 = (m - 1)(m + 1)\)
finding common denominators
Identifying a common denominator is essential for combining fractions. In this exercise, the denominators after factoring are \((m+2)(m-1)\), \((m+1)(m+2)\), and \((m-1)(m+1)\). The least common multiple (LCM) of these denominators is the product of all unique factors: \((m+2)(m-1)(m+1)\). By rewriting each fraction with this common denominator, we obtain:
- \(\frac{m}{(m+2)(m-1)} = \frac{m(m+1)}{(m+2)(m-1)(m+1)}\)
- \(\frac{m}{(m+1)(m+2)} = \frac{m(m-1)}{(m+2)(m-1)(m+1)}\)
- \(\frac{m}{(m-1)(m+1)} = \frac{m(m+2)}{(m+2)(m-1)(m+1)}\)
combining like terms
Once we have a common denominator, we can combine the fractions by adding or subtracting the numerators. In this case:
- \(\frac{m(m+1)}{(m+2)(m-1)(m+1)} = \frac{m(m-1)}{(m+2)(m-1)(m+1)} - \frac{m(m+2)}{(m+2)(m-1)(m+1)}\)
checking solutions
After solving the equation, it's crucial to check the proposed solutions in the original equation. The solutions from the equation \(m(m + 3) = 0\) are \(m = 0\) and \(m = -3\). To validate:
- For \(m = 0\), substitute back into the equation: \( \frac{0}{(0+2)(0-1)} = \frac{0}{(0+3)(0+2)} - \frac{0}{(0-1)(0+1)} \). Each fraction becomes 0, checking out perfectly.
- For \(m = -3\), ensure the denominators are non-zero and verify the balance: \( \frac{-3}{((-3)+2)((-3)-1)} = \frac{-3}{((-3)+3)((-3)+2)} - \frac{-3}{((-3)-1)((-3)+1)} \). Once again, both sides of the equation balance.
