To solve the equation [tex]\(\frac{2m}{m-1} + \frac{m-5}{m^2-1} = 1\)[/tex], we will go through it step by step, making sure to simplify and solve for [tex]\(m\)[/tex]. Here’s the detailed solution:
1. Rewrite the Equation and Factorize:
The equation is:
[tex]\[
\frac{2m}{m-1} + \frac{m-5}{m^2-1} = 1
\][/tex]
Notice that [tex]\( m^2 - 1 \)[/tex] can be factorized:
[tex]\[
m^2 - 1 = (m - 1)(m + 1)
\][/tex]
So, we can rewrite the equation as:
[tex]\[
\frac{2m}{m-1} + \frac{m-5}{(m-1)(m+1)} = 1
\][/tex]
2. Combine the Fractions:
To combine the fractions, we need a common denominator. The common denominator is [tex]\((m-1)(m+1)\)[/tex]. Thus, we rewrite the fractions:
[tex]\[
\frac{2m}{m-1} = \frac{2m(m+1)}{(m-1)(m+1)}
\][/tex]
Now, the equation looks like:
[tex]\[
\frac{2m(m+1)}{(m-1)(m+1)} + \frac{m-5}{(m-1)(m+1)} = 1
\][/tex]
3. Combine the Numerators:
With the common denominator, we can combine the numerators:
[tex]\[
\frac{2m(m+1) + (m-5)}{(m-1)(m+1)} = 1
\][/tex]
Simplify the numerator:
[tex]\[
\frac{2m^2 + 2m + m - 5}{(m-1)(m+1)} = 1
\][/tex]
Combine like terms:
[tex]\[
\frac{2m^2 + 3m - 5}{(m-1)(m+1)} = 1
\][/tex]
4. Cross-Multiplication:
To eliminate the fraction, multiply both sides by [tex]\((m-1)(m+1)\)[/tex]:
[tex]\[
2m^2 + 3m - 5 = (m-1)(m+1)
\][/tex]
5. Expand and Simplify:
Expand the right-hand side:
[tex]\[
2m^2 + 3m - 5 = m^2 - 1
\][/tex]
Move all terms to one side to set the equation to zero:
[tex]\[
2m^2 + 3m - 5 - m^2 + 1 = 0
\][/tex]
Simplify the equation:
[tex]\[
m^2 + 3m - 4 = 0
\][/tex]
6. Solve the Quadratic Equation:
To solve the quadratic [tex]\( m^2 + 3m - 4 = 0 \)[/tex], we can factorize it:
[tex]\[
(m + 4)(m - 1) = 0
\][/tex]
Set each factor to zero:
[tex]\[
m + 4 = 0 \quad \text{or} \quad m - 1 = 0
\][/tex]
Solving for [tex]\(m\)[/tex]:
[tex]\[
m = -4 \quad \text{or} \quad m = 1
\][/tex]
7. Check for Extraneous Solutions:
We must check if these solutions meet the original equation's constraints. Specifically, the values for [tex]\(m\)[/tex] should not make any denominator equal to zero:
- For [tex]\( m = -4 \)[/tex]:
No denominator is zero.
- For [tex]\( m = 1 \)[/tex]:
The denominators [tex]\( (m-1) \)[/tex] and [tex]\((m-1)(m+1)\)[/tex] become zero, which means [tex]\( m = 1 \)[/tex] is not a valid solution because it leads to division by zero.
So, the only valid solution is:
[tex]\[
m = -4
\][/tex]
