Algebra - Exam Question

[martinjr]

I could never understand this question never mind work it out. My problem is the question says 'solve the simultaneous equations' but to me it should read 'equation'. In any case I regarded it as one simultaneous equation but I still could not get a sensible answer. I used substitution for y². Can anyone show me the steps to solve it please:
x²=2y²+y-1,
y²=2x²-11y+8

[petland]

x = 3 y =2
I used the same substitution giving y² = 4y² -9y +6 and then guessed
y = 2 !!!

[Fibonacci]

(1) x²=2y²+y-1,
(2) y²=2x²-11y+8
substitute (1) in (2)
y2= 4y² + 2y - 2 – 11y + 8
simplify
y2 = 4y² -9y +6
subtract y² from both sides
0 = 3y² – 9y +6
Factorise and reverse
(3y – 3)(y-2 ) = 0
So 3y – 3 = 0 or y – 2 = 0
Hence y= 1 or 2
If y = 1 : x² = 2x1² + 1 – 1 = 2, x = sqr 2
If y = 2 : x² = 2x2² +2 – 1 = 9, x = 3
So two solutions
x= sqr 2, y = 1 & x=3 y = 2

[fiction-factory]

There are two equations so the term 'simultaneous equations' is correct. You are looking for the pair/pairs of values of x and y that satisfy both equations at the same time/simultaneously.

If you just take one equation in isolation there would be infinite solutions
eg for x²=2y²+y-1 solutions include
y=0, x= -1
y=1, x= 2
y=2, x= 9
y=-1, x = 0
y=0.5, x=0
.....

But when taken with the second equation there are only 2 pairs of solutions that satisfy both equations simultaneously- as Fibonnaci has shown