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bet you couldn't work this out
Do you think you know the answer to 3X+2-4=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.
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.
That was actually a lot of fun. 
Thats confusing.
Maths has never been my strongest subject
heh... I figured this one out the cheap way
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
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
| wiggy wrote: |
|
Do you think you know the answer to 5X - 3Y = 19 The answer is X = 5 and Y = 2 |
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 co-ordinates 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.
My answer is... ORANGE!
*runs away*
*runs away*
| Lennon wrote: | ||
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 co-ordinates 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. |
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".
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+2-4=4 > 3X+2=4+4 > 3X=4+4-2 > 3X=6 > X=2;
that's right and easy...
Question 3:
5X-3Y=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.
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+2-4=4 > 3X+2=4+4 > 3X=4+4-2 > 3X=6 > X=2;
that's right and easy...
Question 3:
5X-3Y=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.
| Lennon wrote: | ||
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 co-ordinates 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. |
*sigh*
im pretty sure the idea is to watch them stumble, NOT give them the answer.
lol im kidding. yeah youre right.
...It's just maths...
I do equations like these in class... they're not fun.
I do equations like these in class... they're not fun.
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.
It reminds me of how many people struggled through that when I was in middle school, though.
| datter wrote: |
| My answer is... ORANGE!
*runs away* |
I agree with you! Thats totally the correct answer
that seriously is basic alegabra.... really pre alegabra.
| Lennon wrote: | ||
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 co-ordinates 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. |
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.
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.
i knew we could not hav just one ans for that. done so many in school 
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