Solve the system:

y^2 - x^2 = 9

2x - y = 3

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Let `y^2 - x^2 = 9` ..........eq. (i)

`2x - y = 3` .........eq.(ii)

From (ii) we get:

`y=2x-3`

Substituting `y=2x-3` in eq. (i) we get:

`(2x-3)^2-x^2=9`

`rArr 4x^2-12x+9-x^2-9=0`

`rArr 3x^2-12x=0`

`rArr 3x(x-4)=0`

So, either x=0 or x=4

Now plug in the values of x in eq.(ii) to get the corresponding values of y.

So, when x=0 `y=2*0-3=-3`

When x=4 `y=2*4-3=5`

**Therefore, the solutions are (0,-3) and (4,5).**

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