# Math problem - help please

Discussion in 'Secondary Education' started by Upsadaisy, May 15, 2012.

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May 15, 2012

Would you help me solve this?

x/(x + 1) + 3/(x + 4) = (x + 3)/(x + 4)

Thanks so much.

2. 3. ### MathemagicianGroupie

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May 15, 2012

Let's get a common denominator for the LHS! Just like if you had 2/3+1/5, you would have an LCD of 3*5=15, let's take the same approach. Here our LCD is (x+1)(x+4).

So to make the denominator of each (x+1)(x+4), I think of it as multiplying both terms on the left by (x+1)(x+4)/(x+1)(x+4).

So x/(x+1) becomes x(x+4)/(x+1)(x+4) and 3/(x+4) becomes 3(x+1)/(x+1)(x+4).

Now we can combine the like terms on the LHS since the denominator is the same.

So we have

(x*(x+4)+3*(x+1))/(x+1)(x+4)=(x+3)/(x+4).

Now we can multiply both sides by x+4 since we see we can get some nice cancellation.

We have (x*(x+4)+3*(x+1)/(x+1)=x+3.

Now multiply both sides by (x+1), expand, and have a field day:

(x^2+4x+3x+3)=x^2+4x+3

This simplifies to 3x=0.

So x=0.

We plug in, and see that it is indeed correct.

Wolfram Alpha verifies it:

http://www.wolframalpha.com/input/?i=(x/(x+1))+3/(x+4)=(x+3)/(x+4)

Does that make sense? Let me know if something is unclear.

4. ### moparMultitudinous

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May 15, 2012

I was thinking a little differently:

You could subtract the 3/(x+4) to the other side of the equation, so you have:
x/(x+1) = (x+3)/(x+4) - 3/(x+4)

Then combine the two fractions on the right:
x/(x+1) = (x+3-3)/(x+4)

Then I would simplify:
x/(x+1) =x/(x+4)

Cross multiply:

x(x+4) = x(x+1)

Simplify and solve:
x^2 + 4x = x^2 + x

3x=0, so x=0.

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May 15, 2012

That is perfectly clear, Mathemagician. Thank you so much. I guess I just needed a magician to clear things up. I had somehow messed up the denominators along the way... I appreciate your help.

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May 15, 2012

That is really interesting, mopar. Thank you for that insight. Did you jump to that solution right away because of the x + 4 denominator on both sides? I was so fixated on getting a LCD that I didn't even think of that.

7. ### moparMultitudinous

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May 15, 2012

As soon as I saw the x+4 on both sides, I thought it would be much easier than trying to deal with multiply by (x+1)(x+4) to three fractions.

8. ### MathemagicianGroupie

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May 15, 2012

Np, mopar has a much more efficient solution actually! Sometimes with math you can't see the forest through the trees. When I see something like this, I am so quick to find a common denominator that easy tricks slip by me. At least we know both methods work haha!

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May 15, 2012

My student will have to know how to do it the way you presented, though, so I'm glad you did so.

10. ### HMMCohort

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May 15, 2012

mopar...I would do this..

x/(x + 1) = x/(x + 4)

from here x=0 is clearly a solution

so now we can cancel the x's

1/(x + 1) = 1/(x + 4)

reciprocate

x+1=x+4
1=4

thus there an no other solutions. ie the only solution is x=0