# trig

x^2-3y^2=13
x-2y=1
solve for x and y show all your work

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1. wouldn't one put for x in the first equation 2y+1, then solve for y ?

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2. x^2-3y^2=13
x-2y=1
solve for x and y show all your work
Here I substituted the x=2y+1 in as follows:
(2y+1)^2 -3y^2 = 13 =>
4y^2+4y+1 -3y^2 = 13 => cancel out
y^2+4y-12 = 0 => Unfoil
(y+6)(y-2) = 0 =>
Y can = -6 and 2

If that is true then plug those answers into the x-2y = 1 equation as follows:
x-(2)(-6) = 1 =>
x = 1+(2)(-6) =>
x = -11
and
x-(2)(2) = 1 =>
x = 1+(2)(2) =>
x = 5

Y can = -6 and 2
X can = -11 and 5

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