xy=2
{{x^2=3+y^2}}
are you solving for the intersection of the two hyperbolas?
if so, change the first into y = x/2 and sub into the second to get
x^4 - 4x^2 - 12 = 0
(x^2 - 6)(x^2 + 2) = 0
x^2 - 6 or x^2 = -2
x = ±√6 or x = ±√2 i (but we probably only want real intersections)
so y = ±√6 /2
<<if so, change the first into y = x/2 and sub into the second to get >>
should have said:
if so, change the first into y = 2/x and sub into the second to get
no change elsewhere, just a typo
To solve the system of equations xy = 2 and x^2 = 3 + y^2, we can use the method of substitution. Let's go step by step:
Step 1: Start with the equation xy = 2.
Step 2: Solve this equation for x in terms of y by dividing both sides of the equation by y:
x = 2/y.
Step 3: Substitute this value of x in the second equation x^2 = 3 + y^2:
(2/y)^2 = 3 + y^2.
Step 4: Simplify the equation by squaring both the numerator and denominator:
4/y^2 = 3 + y^2.
Step 5: Multiply both sides of the equation by y^2 to eliminate fractions:
4 = 3y^2 + y^4.
Step 6: Rearrange the equation by moving all terms to one side:
y^4 + 3y^2 - 4 = 0.
Step 7: This is a quadratic equation in terms of y^2. Substitute a temporary variable, let's say u = y^2, to simplify the equation:
u^2 + 3u - 4 = 0.
Step 8: Solve this quadratic equation. Factorizing the equation:
(u + 4)(u - 1) = 0.
Step 9: Set each factor equal to zero and solve for u:
u + 4 = 0 or u - 1 = 0.
u = -4 or u = 1.
Step 10: Substitute u back as y^2:
For u = -4, y^2 = -4 (not a real solution for y, as square roots of negative numbers are not real).
For u = 1, y^2 = 1.
So, y = ±1.
Step 11: Substitute the values of y back into the equation x = 2/y to find the corresponding values of x:
For y = 1, x = 2/1 = 2.
For y = -1, x = 2/(-1) = -2.
Step 12: Therefore, the solution to the system of equations xy = 2 and x^2 = 3 + y^2 is:
(x, y) = (2, 1) and (-2, -1).