Discussion in 'Secondary Education' started by Upsadaisy, May 15, 2012.
May 15, 2012
Would you help me solve this?
x/(x + 1) + 3/(x + 4) = (x + 3)/(x + 4)
Thanks so much.
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
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:
This simplifies to 3x=0.
We plug in, and see that it is indeed correct.
Wolfram Alpha verifies it:
Does that make sense? Let me know if something is unclear.
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+4) = x(x+1)
Simplify and solve:
x^2 + 4x = x^2 + x
3x=0, so x=0.
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.
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.
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.
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!
My student will have to know how to do it the way you presented, though, so I'm glad you did so.
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)
thus there an no other solutions. ie the only solution is x=0