ErinWeeks
  • ErinWeeks
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ErinWeeks
  • ErinWeeks
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)}
ErinWeeks
  • ErinWeeks
@ganeshie8 ?
ganeshie8
  • ganeshie8
where is the system

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ErinWeeks
  • ErinWeeks
Sorry lol. x2 - 3y2 = 1 3x2 + 3y2 = 15
ganeshie8
  • ganeshie8
it says use addition method, so its going to be easy.
ganeshie8
  • ganeshie8
oly additions we will be doing..
ErinWeeks
  • ErinWeeks
i figured out its not a .. but i think im making it more harder than it is.
ganeshie8
  • ganeshie8
lets put the equations in row, and ADD them \(vertically\) x2 - 3y2 = 1 3x2 + 3y2 = 15 ----------------- 4x2 + 0 = 16
ganeshie8
  • ganeshie8
4x2 + 0 = 16 x2 = 4 x = \(\pm\) 2
ErinWeeks
  • ErinWeeks
thats kinda were i confused myself at lol
ganeshie8
  • ganeshie8
haa it can be confusing, we have two values for x, x = +2, x = -2
ErinWeeks
  • ErinWeeks
alright and what about 1s though?
ganeshie8
  • ganeshie8
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
ganeshie8
  • ganeshie8
so our solutions are x = \(\pm\)2, y = \(\pm\) 1
ganeshie8
  • ganeshie8
it can be written as :- {(2, 1), (-2, 1), (2, -1), (-2, -1) }
ErinWeeks
  • ErinWeeks
so as C?
ganeshie8
  • ganeshie8
Yes

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