

bet you couldn't work this outwiggy
Do you think you know the answer to 3X+24=4?
The answer is X = 2. Do you think you know the answer to 5X  3Y = 19 The answer is X = 5 and Y = 2 The answer to the poll will be displayed in 7 days. a.Bird
That was actually a lot of fun.
Becky
Thats confusing.
Maths has never been my strongest subject Shanghai_Dragon
heh... I figured this one out the cheap way
Lennon
None of the poll answers are right.
The answer to the poll is: X = Y/7 + 13/7 or put another way: y = 7X  13 Using the following logic. 14X + 5  2Y = 31 14X – 26 = 2Y 7X – 13 = Y X = Y/7 + 13/7 Lennon
This is wrong, you just substitued a few random no's and round 5 for x will give 2 for y. Draw this on a cartesian graph, that is x and y coordinates and you will find it's a linear line with infinite points. You enter any no. for x and you will get a corresponding y value. The line extends to infinity in both axes and both directions with infinite values along this line. datter
My answer is... ORANGE!
*runs away* a.Bird
I got the same answer as you and when I looked at the poll, the answer wasn't up there. What he was trying to ask is "which of the pairs of integers are possible values for x and y". soulman
Question 1:
x=(13+Y)/7 when Y=1, X=2 when Y=2, X=15/7=2.142857... when Y=3 X=16/7=2.28571... ... ... It's impossible to make sure that both X and Y only have 1 value each. Question 2: 3X+24=4 > 3X+2=4+4 > 3X=4+42 > 3X=6 > X=2; that's right and easy... Question 3: 5X3Y=19 > X=(19+3Y)/5 when Y=1, X=21/5=4.5; when Y=2, X=5; when Y=3, X=28/5=5.6; when Y=4, X=6.5; ... ... It's impossible to make sure that both X and Y only have 1 value each. snowboardude
*sigh* im pretty sure the idea is to watch them stumble, NOT give them the answer. lol im kidding. yeah youre right. Nameless
...It's just maths...
I do equations like these in class... they're not fun. eku53ru
I used to have trouble doing the "ax + cy = b" equations unless they were in the "ax + b = cy" format or something. Thanks to the poll, I've found that this problem is now nonexistent (for the most part, anyway).
It reminds me of how many people struggled through that when I was in middle school, though. Becky
I agree with you! Thats totally the correct answer Blaster
that seriously is basic alegabra.... really pre alegabra.
snowboardalliance
Yeah, why did you post some simple math problem that doesn't even make sense? You can't give THE answer to a linear function. As Blaster said, this is really basic pre algebra. Blaster
yea... am i smelling 7th grade over here... wait till you get to high school... and if you are in high school.... then well no comment.
angel_of_death
i knew we could not hav just one ans for that. done so many in school
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