Question 152173: solve the system using your choice of method:
3x^2-2y^2=1
4x-y=3
Answer by orca(409)   (Show Source):
You can put this solution on YOUR website! 3x^2-2y^2=1...........(1)
4x-y=3 ...............(2)
From equation (2) we get
y = 4x - 3.
Substituting it into equation, we have
3x^2 - 2(4x - 3)^2 = 1
Solving for x, we obtain:
3x^2 - 2(16x^2 - 24x + 9) = 1
3x^2 - 32x^2 + 48x - 18 = 1
-29x^2 + 48x - 19 = 0
29x^2 - 48x + 19 = 0
(29x - 19)(x - 1) = 0
So x = 19/29 or x = 1
Substituting x = 19/29 into y = 4x - 3, we have
y = 4(19/29) - 3 = -11/29
Substituting x = 1 into y = 4x - 3, we have
y = 4*1 - 3 = 1
So the solutions to the original simultaneous equations are
x = 19/29
y = -11/29
OR
x = 1
y = 1
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