326. 2y^2 = x^2 - 2 and xy = 2
I solved for y in the second equation and got y = 2/x.
Then I substituted that into the first and got... a bunch of craziness! I know I gotta solve and substitute, but I'm getting lost.. please help
I solved for y in the second equation and got y = 2/x.
Then I substituted that into the first and got... a bunch of craziness! I know I gotta solve and substitute, but I'm getting lost.. please help
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You solved for y for the second equation. Great move. Substitute that for the first equation and solve for x.
2(2/x)² = x² - 2
2(4/x²) = x² - 2
x²[8/x² = x² - 2] [Multiply both sides by x²]
8 = x^4 - 2x²
0 = x^4 - 2x² - 8 [Rearrange the terms]
0 = (x² - 4)(x² + 2)
x² - 4 = 0 and x² + 2 = 0
x² = 4 and x² = -2
x = ±2 [No solutions for x² + 2 = 0 and they are imaginary; ±i√2. Also; there are no real numbers for determining x² = -2]
Finally, substitute each of the x values for the second equation and solve for y.
When x = 2:
y = 2/2
y = 1
When x = -2:
y = 2/-2
y = -1
Therefore, the solution are (2,1) and (-2, -1)
I hope this helps!
2(2/x)² = x² - 2
2(4/x²) = x² - 2
x²[8/x² = x² - 2] [Multiply both sides by x²]
8 = x^4 - 2x²
0 = x^4 - 2x² - 8 [Rearrange the terms]
0 = (x² - 4)(x² + 2)
x² - 4 = 0 and x² + 2 = 0
x² = 4 and x² = -2
x = ±2 [No solutions for x² + 2 = 0 and they are imaginary; ±i√2. Also; there are no real numbers for determining x² = -2]
Finally, substitute each of the x values for the second equation and solve for y.
When x = 2:
y = 2/2
y = 1
When x = -2:
y = 2/-2
y = -1
Therefore, the solution are (2,1) and (-2, -1)
I hope this helps!