simultaneous equation one linear and one quadratic

Question

25p^2-7q^2=29 and 5p+7q=-1                                                 

Answer 1

25p^2 - 7q^2 = 29
25p^2 = 7q^2 + 29

5p + 7q = -1
5p = –7q – 1
(5p)^2 = (–7q – 1)^2
25p^2 = 49q^2 + 14q + 1

49q^2 + 14q + 1 = 7q^2 + 29
42q^2 + 14q – 28 = 0
3q^2 + q – 2 = 0
(3q – 2)(q + 1) = 0
q = –1 or q = 2/3

5p = –7(–1) – 1 = 6 ==> p = 6/5
5p = –7(2/3) – 1 = –17/3 ==> p = –17/15

Solutions are:
(p,q) = (6/5,–1) and (–17/15,2/3)

check:

25(6/5)^2 - 7(–1)^2 =? 29
5(6/5) + 7(–1) =? –1

36 – 7 = 29 √
6 – 7 = –1 √, so (6/5,–1) IS a solution

25(–17/15)^2 - 7(2/3)^2 =? 29
5(–17/15) + 7(2/3) =? –1

17^2/9 – 28/9 =? 29
–17/3 + 14/3 =? –1

289/9 – 28/9 =? 29
–17/3 + 14/3 =? –1

261/9 = 29 √, so (–17/15,2/3) IS a solution
–3/3 = –1 √

See http://www.wyzant.com/resources/files/266305/line_hyperbola_intersection

Answer 2

The most straightforward way is probably to solve for p or q in the linear equation and then substitute into the first equation.

p = (-1-7q )/5

 

(-1-7q)²-7q²= 29

49q²+14q+1-7q²=29

42q²+14q-28 = 0

 

That looks like an equation you can deal with. Once you find q, find the corresponding p using the linear equation.

You should get two ordered pairs as your answer.