Question

I Tried Doing This On My Own & I Figured Out Its Not A But I Cant Get Any Further..
Solve the system by the addition method.

Answer 1

A. {( 1, 2), ( -1, 2), ( 1, -2), ( -1, -2)}

B. {( 1, 2), ( -1, -2)}

C. {( 2, 1), ( -2, 1), ( 2, -1), ( -2, -1)}

D. {( 2, 1), ( -2, -1)}

Answer 2

@ganeshie8 ?

Answer 3

where is the system

Answer 4

Sorry lol. x2 - 3y2 = 1
3x2 + 3y2 = 15

Answer 5

it says use addition method, so its going to be easy.

Answer 6

oly additions we will be doing..

Answer 7

i figured out its not a .. but i think im making it more harder than it is.

Answer 8

lets put the equations in row, and ADD them \(vertically\)

x2 - 3y2 = 1
3x2 + 3y2 = 15
-----------------
4x2 + 0 = 16

Answer 9

4x2 + 0 = 16

x2 = 4

x = \(\pm\) 2

Answer 10

thats kinda were i confused myself at lol

Answer 11

haa it can be confusing,
we have two values for x,
x = +2, x = -2

Answer 12

alright and what about 1s though?

Answer 13

lets solve y, when x = +2,

put x = 2 in first equation,
x2 - 3y2 = 1

2^2 - 3y2 = 1

4 - 3y2 = 1
3 = 3y2
y = \(\pm\) 1

Answer 14

so our solutions are x = \(\pm\)2, y = \(\pm\) 1

Answer 15

it can be written as :-

{(2, 1), (-2, 1), (2, -1), (-2, -1) }

Answer 16

so as C?

Answer 17

Yes