# Math Help - Wonder if anyone can help with something raised in further maths AS today?

1. ## Wonder if anyone can help with something raised in further maths AS today?

I was in further maths today, and came across an interesting situation within adding fractions. I wondered if there was any way you could have 2 different fractions a/b + c/d =a+c/bd where a,b,c and d are all integers. I would appreciate a response soon.
Many thanks

2. Do you mean when can (a/b) + (c/d) = (a+c)/(bd) happen? You have to be very careful with parentheses when writing expressions like this. Technically, you wrote

(a/b) + (c/d) = a + (c/bd), which is very different.

3. yep. I am kind of using general notation on paper, but indeed, it would be (a/b)+(c/d)=(a+c)/(bd). bad habit, should probably get used to doing that the right way

4. Ok. You'd need to set your expression equal to the expression that is always true. That is, you set

$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}=\dfrac {a+c}{bd}.$

The expression in the middle is always true, and the expression on the far right is your expression. Now compare the two. When could they be equal?

5. This was the problem we had. We are a class of 6, and our teacher. We came across the problem in the lesson, and wondered if it were possible for this to happen. I assume there must be a point when this CAN happen.

6. My post # 4 was intended as a sort of prod to get you to see how to do the next step in your problem. The situation you describe can indeed happen. However, it is not the philosophy of this forum (nor me) to simply hand out solutions to problems. We're here to help people get unstuck with problems and then solve the rest of the problem themselves. It's much better for the student that way, because then you own the solution. You understand the solution. If I just tell a student the answer, much data shows that the likelihood of the student remembering is quite small. Whereas, if the student solves the problem by himself/herself, they will own the solution and remember the thought process by which they arrived at the solution much better. So take another look at post # 4, and see if you can see where to go next.

7. On this information, I have had a though about this. If 2a+c=0
Therefore, a=7, b=3, c=-14, d=5
(7/3)+(-14/5)=(35-43)/(3*5)=-7/15
Likewise, 7+-14=-7
3*5=15,
Thus, (a+c)/bd=((ad)+(bc))/bd

8. Your arithmetic is incorrect. Check out your added fractions again. This approach, incidentally, may eventually give you a particular case that works. But you would not have found all situations in which it can happen. In fact, I think there's only one equation with four unknowns that characterizes the situations in which this can occur. So you'd get a 3-parameter family of solutions. That is, there are infinitely many possible ways to make this work.

9. I believe that the added fractions are correct. However, if you know of any problem with them, could I please be shown the correction??

10. $(7/3)+(-14/5)\not=(35-43)/(3*5)$.

You found the common denominator, which is 15. However, $-14\times 3\not=-43.$ You know this can't be correct, because on the LHS side you have an even integer times an odd integer, which is always even, and yet the RHS is odd. So what is $-14\times 3?$