Question

Which shows solutions to x + 3y = 7 if x and y must be whole numbers
{(0, 5), (1, 3), (2, 1)}
{(2, 1), (1, 4), (0, 7)}
{(1, 2), (4, 1), (7, 0)}
{(0, 5), (1, 4), (2, 1)}

Answer 1

Plug each pair in for x and y and see which ones work...ie which ones give you 7.
In each pair (__,__) plug the first number in for x and the 2nd for y

Answer 2

plug each point into the equation and see if you get a true result

for instance, plug in (0,5) to get


x + 3y = 7

0 + 3*5 = 7 .. plug in x = 0, y = 5

0 + 15 = 7

15 = 7 ... this is FALSE

so because the last equation is false, this means that (0,5) is NOT a solution to x+3y = 7

Answer 3

ok so this answer would be the {(0, 5), (1, 4), (2, 1)}??

Answer 4

how can the answer be that if I just proved that (0,5) is not a solution to x+3y = 7

Answer 5

oh sph sorry my mind isblank right now

Answer 6

what if I think the answer is #2 is that right

Answer 7

pick any point from that set and plug it into the equation

Answer 8

if you get a true result, then check the other points

if you get a false result, then you can eliminate that option

Answer 9

well can you just tell me if I got that answer right? #2?

Answer 10

let's plug in (2,1)

x + 3y = 7

2 + 3*1 = 7

2 + 3 = 7

5 = 7

that's false, so (2,1) is NOT a solution

Answer 11

see how this works?

Answer 12

lol kinda of ok well what about #1

Answer 13

(0,5) is part of choice A, so what do you think?

Answer 14

remember above (0,5) is shown to not be a solution to x+3y = 7

Answer 15

nope so I imam say its 3 right

Answer 16

choice C is the only thing left, but I recommend you check all the points to make sure

Answer 17

what this is a mid untit test I needa know is it right...

Answer 18

you can't ask for help on tests