Question

PLEASE HELP!!!!!!

Answer 1

@hartnn @ganeshie8

Answer 2

where is the system ?

Answer 3

Sorry, this is the question

Answer 4

you'll have to check each and every option.
like
1st option :
add 1st equation in sys1 (4x-5y=2) to 2 times 2nd equation 2(3x-y=8)

see if you get 3x-8y =4

Answer 5

THe ans is choice B...system 2 and system 3..!!!

Answer 6

Thanks @zaibali.qasmi :), but i kinda wanted 2 work out the problem

Answer 7

For the first option I got 11x - 11y = 12. So that's wrong, right?

Answer 8

good.
now go for 2nd option

Answer 9

bdw,
(4x-5y=2)+ 2(3x-y=8)

4x+6x -5y -2y =2+16

10x -7y = 18
which is not 3x-8y = 4
hence its not 1st option

Answer 10

Ok, so for option 2, I added 4x-5y = 2 twice into 3x-8y=4 and got 11x-18y=8

Answer 11

So that isn't right?

Answer 12

1st equation in sys 2 is 4x-5y =2
2 times 2nd equation in sys 2 is 2 (3x-8y= 4)
adding them

\((4x-5y=2) + (6x-16y=8)\)

do you get that as 10x -21y =10
?

Answer 13

how did you get
11x-18y=8

if you show your work, i will help you spot the error :)
or try again

Answer 14

Oooh, I get it :)

Answer 15

I thought I was supposed 2 add the first equation twice, not multiply the second equation by 2 XD

Answer 16

Thanks :) <3

Answer 17

welcome ^_^