+1993 
im unsure how to go forth in answering this question, can someone help me?
+128475 Second one, first [easier to see ]
x^2 + y^2 = 9 (1)
x^2 y^2
___ - _____ = 1 (2)
9 4
Multiply (2) through by 4 and we get that
(4/9) x^2 - y^2 = 4 (3)
Using (1) and (3) we have that
x^2 + y^2 = 9
(4/9)x^2 - y^2 = 4 add these equations
(13/9)x^2 = 13 multiply both sides by (9/13)
x^2 = 9 take both roots
x = -3 or x = 3
Using (1) ....no matter if x = -3 or x = 3.....we have that
(3)^2 + y^2 = 9
9 + y^2 = 9 subtract 9 from both sides
y^2 = 0
y = 0
So....the solutions are (3, 0) and (-3,0)
The two graphs are a circle and a hyperbola : https://www.desmos.com/calculator/wwigbxzcsz
+1993 I will make note of this ! Thanks for explaining it piece by piece, it helps me understand!
+128475 y^2 x^2
___ - _____ = 1 (1)
4 2
y^2 = 8(x + 1) (2)
Sub (2) into (1) and we get that
8(x + 1) x^2
_______ - _____ = 1 Multiply this through by 4
4 2
8(x + 1) - 2x^2 = 4
8x + 8 - 2x^2 = 4 rearrange as
2x^2 - 8x - 4 = 0 divide through by 2
x^2 - 4x - 2 = 0
Using the quad formula
x = 4 ±√[ 4^2 - 4 (1) (-2) ] 4 ±√24 4 ±√[4*6] 4 ±2√6
______________________ = ________ = _________ = ______ = 2±√6
2 * 1 2 2 2
When x = 2 + √6 then we have that
y^2 = 8( 2 + √6 + 1)
y^2 = 8 ( 3 + √6) take both roots
y = ±√[ 8 (3 + √6) ]
y = ±2√[ 2 (3 + √6 ]
y = ±2√[6 + 2√6 ]
And when x = 2 -√6 we have that
y^2 = 8( 2 - √6 + 1) take both roots
y = ±√[ 8 ( 3 - √6) ]
y = ±2√[2(3 - √6) ]
y = ±2√[6 - 2√6 ]
So......
When x = 2 +√6....then y = 2√[6 + 2√6 ] and -2√[6 + 2√6 ]
When x = 2 - √6....then y = 2√[6 - 2√6 ] and -2√[6 - 2√6 ]
+1993 hmm, I kind of confused still. I tried looking at your work and then comparing it to the answer choices, but im unsure which ones it would be? these two?
+128475 The correct choices are
(2 + √6 , 2√[ 6 + 2√6 ] )
(2 + √6 , -2√[ 6 + 2√6 ] )
(2 - √6 , 2√[ 6 - 2√6 ] )
(2 - √6 , -2√[ 6 - 2√6 ] )
+1993 ohhh okay, I see the answers now . It all makes sense ! Thank you for helping me with these 2 questions!
