Algebraic fractions in a linear sequence: 2/x+3, (missing term), 2/2x-3

[Simonsky]

Hi Folks,

Came across this question which I can't work out and keep coming up with wrong answer:

Three terms of a linear sequence-you have to find the missing middle term:

2/x+3, (missing term), 2/2x-3

So I tried it like this: I thought if I take the first term away from the third I get 2x the difference then half that and add it on to the first term to get the middle one - but I think there must be a better way that I can't think of as I kept getting the wrong answer using that method.

Any help appreciated as always.

 

[HallsofIvy]

Hi Folks,

Came across this question which I can't work out and keep coming up with wrong answer:

Three terms of a linear sequence-you have to find the missing middle term:

2/x+3, (missing term), 2/2x-3

So I tried it like this: I thought if I take the first term away from the third I get 2x the difference then half that and add it on to the first term to get the middle one - but I think there must be a better way that I can't think of as I kept getting the wrong answer using that method.

Any help appreciated as always.

What you say is perfectly good. How do you know you "kept getting the wrong answer using that method"? (One complication is that what you wrote would be properly interpreted (2/x)+ 3 and (2/2x)- 3 but I feel sure you meant 2/(x+ 3) and 2/(2x- 3).)

The difference between 2/(x+ 3) and 2/(2x- 3) is [2(x+ 3)- 2(2x- 3)]/((x+3)(2x- 3))= (-2x+ 12)/((x+3)(2x- 3)). Half of that is (-x+ 6)/((x+ 3)(2x- 3)). Adding that to 2/(x+ 3), 2/(x+ 3)+ (-x+ 6)/((x+ 3)(2x- 3))= (2(2x- 3)- x+ 6)/((x+ 3)(2x- 3)= 3x/((x+3)(2x- 3)).

Simpler, but giving the same answer, is to average the two. 2/(x+ 3)+ 2/(2x- 3)= (2(2x- 3)+ 2(x+ 3))/((x+ 3)(2x- 3)= 6x/((x+ 3)(2x- 3)) and then dividing by 2, 3x/((x+ 3)(2x- 3)).

In general, if a and b are two numbers, the distance between them is b- a. Half of that is (b- a)/2 and adding to a, the midpoint is a+ (b- a)/2= (2a+ b- a)/2= (a+ b)/2, the average of the two numbers.

 

[Simonsky]

Thanks -I think what I did was instead of halving the numerator I multiplied by 1/2 so doubled the denominator - thanks for that!

 

[HallsofIvy]

Thanks -I think what I did was instead of halving the numerator I multiplied by 1/2 so doubled the denominator - thanks for that!

?? Doubling the denominator is exactly the same as halving the numerator. That shouldn't have caused any problem.

 

[Dr.Peterson]

Thanks -I think what I did was instead of halving the numerator I multiplied by 1/2 so doubled the denominator - thanks for that!

Can you show us what answer you got, and why you think it is wrong? It may be that it was called wrong just because it was not simplified, and simplification was required.