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SOLVE THE NONLINEAR SYSTEM OF EQUATIONS FOR REAL SOLUTIONS
 
x2 + 3y2 = 6
 
x - 3Y2 =  10

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Ron S. | Expert in Test Prep, Science, and MathExpert in Test Prep, Science, and Math
4.8 4.8 (51 lesson ratings) (51)
1
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!
 
Kathye P. | Math Geek, passionate about teachingMath Geek, passionate about teaching
5.0 5.0 (150 lesson ratings) (150)
0
Hi, Mario.
 
As another tutor pointed out, if you graphed these equations, the figures would not intersect. You can also determine this algebraically.
 
x2 + 3y2 = 6

x2 - 3y2 = 10
 
Adding the 2 equations causes the y terms to cancel, leaving:
 
2x2  = 16
 x2   = 8
 x  = ±√8
 
Plugging that back into the first equation  to solve for y, we get:
 
8 + 3y2 = 6
 
3y2 = -2
 
y2 = - 2/3
 
-2/3 has no real square root, so there is no solution to the system. 
 
Hope this helps!
Kathye P.