SOLVE THE NONLINEAR SYSTEM OF EQUATIONS FOR REAL SOLUTIONS

x

^{2}+ 3y^{2 = }6x

^{2 }- 3Y^{2 = }10-
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SOLVE THE NONLINEAR SYSTEM OF EQUATIONS FOR REAL SOLUTIONS

x^{2} + 3y^{2 = }6

x^{2 } - 3Y^{2 = }10

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There are no real solutions to this system; the first equation is an ellipse whose x-intercepts are ±√6, while the second is a hyperbola whose x-intercepts are ±√10, so there no intersection.

If the two equations had the same constant on the right hand side, there would be two solutions, equal to the square root of that constant.

Hope this helps!

Hi, Mario.

As another tutor pointed out, if you graphed these equations, the figures would not intersect. You can also determine this algebraically.

x^{2} + 3y^{2} = 6

x^{2} - 3y^{2} = 10

x

Adding the 2 equations causes the y terms to cancel, leaving:

2x^{2} = 16

x^{2} = 8

x = ±√8

Plugging that back into the first equation to solve for y, we get:

8 + 3y^{2} = 6

3y^{2} = -2

y^{2} = - 2/3

-2/3 has no real square root, so there is no solution to the system.

Hope this helps!

Kathye P.

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