Fractional Equation Help: Solve 3 Problems

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SUMMARY

The forum discussion centers on solving three fractional equations: \( \frac{x}{x-1} - 1 = \frac{3}{x+1} \), \( \frac{4}{b} - \frac{1}{b+3} = \frac{3b+2}{b^2+2b-3} \), and \( \frac{3r+1}{r+3} + 2 = \frac{5r-2}{r+3} \). The first equation can be simplified by multiplying through by \( (x-1) \) and \( (x+1) \) to eliminate fractions, leading to \( x(x+1) + (x-1)(x+1) = 3(x-1) \). The discussion emphasizes the importance of careful manipulation of fractions and parentheses in solving these types of problems.

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There are three problems on my homework I can't get quite right.

x/x-1 - 1 = 3/x+1

4/b - 1/b+3 = 3b+2/b^2+2b-3

3r+1/r+3 + 2 =5r-2/r+3
 
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Mikeybr said:
There are three problems on my homework I can't get quite right.

x/x-1 - 1 = 3/x+1

4/b - 1/b+3 = 3b+2/b^2+2b-3

3r+1/r+3 + 2 =5r-2/r+3

Hi Mikeybr. Welcome to MHB! :)

Try posting one question at a time please and show some attempt at it. This helps us help you more efficiently. Also, be careful with parentheses!

I believe #1 is the following:

[math]\frac{x}{x-1}-1=\frac{3}{x+1}[/math]

You could do this a few ways but the idea is to get rid of the fractions somehow. Let's first try to rewrite the first fraction by multiplying everything by $(x-1)$.

[math](x-1) \cdot \frac{x}{x-1} -1 \cdot (x-1) = (x-1) \cdot \frac{3}{x+1}[/math]

The first fraction now has [math]\frac{x-1}{x-1}[/math] in it, which is 1 so that fraction is now just x. The rest of the equation is below.

[math]x - (x-1) = \frac{3(x-1)}{x+1}[/math]

Now we want to change the other fraction, so we can multiply everything by $(x+1)$. Doing that we get:

[math]x(x+1)+(x-1)(x+1)=3(x-1)[/math]

Can you solve from here?
 

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