Question

Prove or disprove: The set {(x, y),(a, b) | 3a + b = x, y = 2x − a}

is a function
from R × R to R × R.

Answer 1

@jim_thompson5910

Answer 2

so the rule is (x,y) ----> (a,b) where 3a + b = x, y = 2x − a

Answer 3

Would you begin by showing that a,b are arbitrary elements in RxR
thus since x = 3a +b and y = 2(3a+b) - a then x,y must also be in RxR?

Answer 4

so (a,b) is the input, (x,y) is the output?

Answer 5

I'm not sure which is which

Answer 6

Other way around, sorry.
(a,b) is input

Answer 7

so if this was a function, then (a,b) would map to EXACTLY ONE ordered pair (x,y)

Answer 8

correct

Answer 9

if (a,b) produced more than one ordered pair, then you can use that to disprove it's a function

Answer 10

Correct, I cannot find any (a,b) that produces more than a single ordered pair, so how would I being proving this to be true?

Answer 11

let's focus on 3a + b = x first

Answer 12

3a + b is linear for a and b, so if you imagine setting up a 3-space axis (x, y, z axis), then 3a+b would be a plane in this 3-space setting

Answer 13

for any given a and b, you get some height x, which is exactly one height I think

Answer 14

correct

Answer 15

so that leads me to think that 3a+b = x produces one value of x given any fixed (a,b) value

Answer 16

Could you simply write (a,b) maps to (3a +b, 5a -b)?

Answer 17

Could you simply write (a,b) maps to (3a +b, 5a + b)? rather

Answer 18

how did you get the second value in the ordered pair?

Answer 19

oh you plugged in x = 3a+b

Answer 20

into y = 2x − a and simplified

Answer 21

y = 2x - a, thus y = 2(3a +b) - a thus y = 6a + b - a, which means y = 5a + b

Answer 22

i gotcha

Answer 23

yeah a --> 3a+b which is basically mapping a line to a line

b ---> 5a+b is also a line-to-line mapping

Answer 24

so I think that's enough info to say that any given input produces one output

Answer 25

and since they are linear there is only one input/output pair thus it must be a function by the horizontal line test?

Answer 26

you mean vertical line test, but yes that's correct

Answer 27

Yea sorry, my bad.
Thank you for your help.

Answer 28

you're welcome