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Yunie
Sep 2nd, 2007, 07:12 AM
Given that (x+37/5) and (7x-1/2) are two consecutive even integers, find the value of x.
Ans: x =3
Hey all, I don't understand on how to do and solve this question. Could someone help me? Thanks a lot, I really appreciate your kind help. Thanks. :)
zaza
Sep 2nd, 2007, 08:43 AM
Can you show us how far you have gotten? I think it is time you started to work some of these questions through...it will definitely be better for your understanding that way :)
Yunie
Sep 2nd, 2007, 09:09 PM
The problem is that I don't understand the question...But then, I have thought of adding up the (x+37/5) and (7x-1/2) together...Is that correct?
Yunie
Sep 3rd, 2007, 01:30 AM
Help please? Thanks...:)
Thomas154321
Sep 3rd, 2007, 06:22 AM
You need to use the fact that they are consecutive. Start with the first number, add 2 and you get the second. But you don't know which is the smaller number so do it both ways. One works the other doesn't.
opus
Sep 3rd, 2007, 07:16 AM
?? Maybe I have a mindblock, but i can't find an answer that would make any of the given terms to an integer????
I did understand (x+37/5) and (7x-1/2) as in (x+7.4) and (7x - 0.5).
Thomas154321
Sep 3rd, 2007, 10:58 AM
The integers are actually [(x+37)/5] and [(7x - 1)/2]. I found the same answer as you first time around too.
opus
Sep 3rd, 2007, 03:34 PM
I hate it when those kids can't even copy&paste correctly.
Code Doc
Sep 12th, 2007, 05:56 PM
Let
Y1 = (x + 37)/5
Y2 = (7x - 1)/2
"Two consecutive even integers" means that (Y1 + Y2)/2 = Y1 + 1
Therefore, [(x + 37)/5 + (7x - 1)/2]/2 = [(x + 37)/5] + 1
Solve for x. OP blew the original problem by leaving out parentheses. As stated, there was no solution.:sick:
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