solve the system
7x^2 - 5y^2 + 20y = 3
21x^2 + 5y^2 =209
since 7x^2 = 5y^2 - 20y + 3
21x^2 = 15y^2 - 60y + 9
plug that into the 2nd equation to get
20y^2 - 60y - 200 = 0
y^2 - 3y - 10 = 0
(y-5)(y+2) = 0
From there you can see that the circle and the hyperbola intersect at
(3,-2) (-3,-2) (2,5) (-2,5)
To solve the system of equations:
Step 1: Rearrange the first equation to isolate the variable x:
7x^2 - 5y^2 + 20y = 3
7x^2 = 5y^2 - 20y + 3
Step 2: Simplify and rewrite the second equation:
21x^2 + 5y^2 = 209
Step 3: Substitute the expression obtained in Step 1 into the second equation:
21(5y^2 - 20y + 3) + 5y^2 = 209
Step 4: Distribute and simplify:
105y^2 - 420y + 63 + 5y^2 = 209
110y^2 - 420y + 63 = 209
Step 5: Move all terms to one side to obtain a quadratic equation:
110y^2 - 420y + 63 - 209 = 0
110y^2 - 420y - 146 = 0
Step 6: Solve the quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's use the quadratic formula:
The quadratic formula is given by:
y = (-b ± √(b^2 - 4ac)) / (2a)
In this equation, a = 110, b = -420, and c = -146.
Plugging these values into the quadratic formula gives:
y = (-(-420) ± √((-420)^2 - 4 * 110 * (-146))) / (2 * 110)
Simplifying further:
y = (420 ± √(176400 + 64240)) / 220
y = (420 ± √(240640)) / 220
y = (420 ± 490.56) / 220
Step 7: Solve for y by considering both the positive and negative values:
Case 1: y = (420 + 490.56) / 220
y = 910.56 / 220
y ≈ 4.14
Case 2: y = (420 - 490.56) / 220
y = -70.56 / 220
y ≈ -0.32
Step 8: Substitute the obtained y-values back into the first equation to solve for x.
For case 1 (y ≈ 4.14):
7x^2 - 5(4.14)^2 + 20(4.14) = 3
7x^2 - 107.73 + 82.8 = 3
7x^2 - 24.93 = 3
7x^2 = 27.93
x^2 = 27.93/7
x^2 = 3.99
x ≈ ±1.99
For case 2 (y ≈ -0.32):
7x^2 - 5(-0.32)^2 + 20(-0.32) = 3
7x^2 - 0.512 + (-6.4) = 3
7x^2 - 6.912 = 3
7x^2 = 9.912
x^2 = 9.912/7
x^2 = 1.416
x ≈ ±1.19
Therefore, the system of equations has four approximate solutions:
(x ≈ 1.99, y ≈ 4.14), (x ≈ -1.99, y ≈ 4.14), (x ≈ 1.19, y ≈ -0.32), (x ≈ -1.19, y ≈ -0.32).