Solve the system:

x^2 - y = 1

2x^2 + y^2 = 17

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`x^2-y = 1` -----(1)

`2x^2+y^2 = 17` ---(2)

From (1)

`x^2 = 1+y`

Substituting this in (2) yields;

`2(1+y)+y^2 = 17`

`y^2+2y-15 = 0`

`y^2+5y-3y-15 = 0`

`y(y+5)-3(y+5) = 0`

`(y+5)(y-3) = 0`

`y = -5 or y = 3`

From (1)

`x^2 = 1+y`

When y = -5

`x^2 = -4` which x has no real solutions.

When `y = 3`

`x^2 = 4`

`x = +-2`

*So the answers are x = 2,y = 3 and x = -2, y = 3*

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