The first equation is a hyperbola, while the second is an ellipse.

We multiply the first row by 4 so we can eliminate the `y^2` term:

`12x^2-4y^2=16`

`x^2+4y^2=10`

Now adding the two rows, we obtain:

`13x^2=26`

`x^2=2`

`x=+-sqrt(2)`

Substituting `+sqrt(2)` into the question's first equation gives us

`y=+-sqrt(2)`

Likewise, substituting `-sqrt(2)` into the first equation also gives us

`y=+-sqrt(2)`

This gives us the solutions

`(sqrt(2),sqrt(2)),` `(sqrt(2),-sqrt(2)),` `(-sqrt(2),sqrt(2)),` `(-sqrt(2),-sqrt(2)) `

≈ (±1.414, ±1.414), (±1.414, ∓1.414)

The graph shows the intersection of the ellipse and the hyperbola. We see 4 intersection points, with the same values that we found algebraically.

12345-1-2-3-4-5123-1-2-3xyOpen image in a new page

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