Adding fractions

Algebra Level 1

True or False :

If a , b , c , and d a, b, c,\text{ and } d are positive integers, then

a b + c d = a + c b + d . \frac{a}{b} + \frac{c}{d} = \frac{a+c}{b+d}.

False True

Zandra Vinegar by Zandra Vinegar

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6 solutions

a b + c d = a d + b c b d ≠ a + b c + d \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\neq \frac{a+b}{c+d}

31 Helpful 0 Interesting 0 Brilliant 1 Confused
Michael Wiley
Sep 29, 2016

Is 1/2 + 1/4 = 2/6! Of course not!

13 Helpful 0 Interesting 0 Brilliant 0 Confused

transpose and cross multi.

Roger Worrall - 4 years, 1 month ago
Raushan Sharma
Mar 23, 2016

This is true only when a b = c d \frac{a}{b} = \frac{c}{d} , which is known as A d d e n d o Addendo p r o p e r t y property .

12 Helpful 0 Interesting 0 Brilliant 0 Confused

It is false always. Because if a/b = c/d , then LHS is twice of RHS.

Sahil Jain - 4 years, 2 months ago

Rushan - please remove your post. This is NOT the addendo property. The addendo property is if a/b = c/d = (a+c)/(b+d). This problem is a/b + c/d = (a+c)/(b+d) - notice the + between a/b and c/d. In the addendo property there is an = sign.

Drue Gawel - 4 years, 1 month ago

Raushan - Wrong

if a/b = c/d = 1/2 a/b + c/d = 1/2 + 1/2 = 1 Actual

But as per above formula a/b +c/d = (a+c)/(b+d) = 1+1 /2+2 = 2/4 = 1/2

lakhi kukreja - 3 years, 4 months ago

a b + c d = a d + b c b d ≠ a + c b + d \frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}\neq \frac{a+c}{b+d}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Everyone knows that you don't add up the denominator's together. It is false!

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Ahmad Naji
Jan 15, 2016

Are you kidding me... no solutions needed 😂😂😂

0 Helpful 0 Interesting 0 Brilliant 0 Confused

please do not comment again with 'are you kidding me' its rude and I'm doing this trying to learn algebra

Ray Soto - 4 years, 2 months ago

let us answer on the question

iraj honardan - 4 years, 1 month ago

Firstly, I agree with Ray Soto in that all the commentors could be of different age, different mathmatical abilities and / or at different stages of their learning ! ( We've all got to start somewhere ! Ahmad, for what its worth, I don't think you were being deliberatefully hurtful and that was just a flippant, throwaway remark ! My opinion is that the equation could be made to work by varying the values of the integers to fit *! I suppose that you could even have 2 different letters having the same value ? My school days are a long way ago, about 40 yrs, and, if I remember correctly, that would not be the case as it is not the way or befitting the reasoning of algebra, working on different values for different numbers to distinguish between unknown ( at least at the outset ) different values for each number ! Could anyone out there explain if that would be possible ? I excersised a few practical examples and it came out FALSE !

Stephen Elliott - 4 years, 1 month ago

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