12x² - 5y² = 7 ①

Sub x = 1 and y = p,

12(1²) - 5p² = 7
12 - 5p² = 7 ③

2p²x - 5y = 7 ②

Sub x = 1 and y = p,

2p²(1) - 5p = 7
2p² - 5p = 7 ④

③ = ④ since both have the right hand side equalling 7.


12 - 5p² = 2p² - 5p
7p² - 5p - 12 = 0
(7p - 12)(p + 1) = 0
7p - 12 = 0 or p + 1 = 0
7p = 12 or p = -1
p = 12/7 (rejected as it does not satisfy both equations ③ and ④)

Next, since we know p = -1, then p² = (-1)² = 1

We can sub p² = 1 into ②

12x² - 5y² = 7 ①
2x - 5y = 7 ②

From ②,
2x = 5y + 7

Square both sides,

(2x)² = (5y + 7)²
4x² = 25y² + 70y + 49

Multiply throughout by 3,

12x² = 75y² + 210y + 147

Sub into ①,

75y² + 210y + 147 - 5y² = 7
70y² + 210y + 140 = 0

Divide throughout by 70,
y² + 3y + 2 = 0
(y + 1)(y + 2) = 0
y = -1 or y = -2

Clearly, y = -2 is part of the other solution. Sub back into the part from ②

2x = 5(-2) + 7
2x = -3
x = -1.5

∴ (-1.5, -2) is the other solution

Slightly different way to find p : solve for p using separately for each equation first.

12x² - 5y² = 7 ①

Sub x = 1 and y = p,

12(1²) - 5p² = 7
12 - 5p² = 7
5p² = 5
p² = 1
p = ±√1
p = 1 or -1

Then,
2p²x - 5y = 7 ②

Sub x = 1 and y = p,

2p²(1) - 5p = 7
2p² - 5p - 7 = 0
(2p - 7)(p + 1) = 0
2p = 7 or p = -1
p = 3.5


Since the value of p must satisfy both equations, p = -1