Question 1182579
.
Solve the system by using Elimination Method:
1. 3x² + y² = 21
4x² - 2y² = -2
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Your starting equations are

    3x^2 +  y^2 = 21      (1)

    4x^2 - 2y^2 = -2      (2)



Multiply equation (1) by 2 (both sides).   Keep equation (2) as is

    6x^2 + 2y^2 = 42      (1')

    4x^2 - 2y^2 = -2      (2')


Now add equations (1') and (2').  The terms " 2y^2 "  and  " -2y^2 "  will cancel each other (elimination),
and you will get single equation for one unknown x, only:


    6x^2 + 4x^2 = 42 + (-2)

        10x^2   = 40

          x^2   = 40/10 = 4

          x             = {{{sqrt(4)}}} = +/- 2.


For now, we have two solutions for x:  +2  and -2.


Substitute x= 2 into equation (1).  You wil get then

    3*2^2 + y^2 = 21  --->  3*4 + y^2 = 21  --->  y^2 = 21 - 12 = 9  --->  y = {{{sqrt(9)}}} = +/- 3

So, with x= 2, you have two solutions  (x,y) = (2,3)  and  (x,y) = (2,-3).



Next, substitute x= -2 into equation (1).  You wil get then

    3*(-2)^2 + y^2 = 21  --->  3*4 + y^2 = 21  --->  y^2 = 21 - 12 = 9  --->  y = {{{sqrt(9)}}} = +/- 3.

So, with x= -2, you have two solutions  (x,y) = (-2,3)  and  (x,y) = (-2,-3).


Thus you get 4 (four) different pairs solutions for the given equations.


<U>ANSWER</U>.  The given system has 4 (four) solutions  (x,y) = (2,3), (2,-3), (-2,3)  and  (-2,-3).
</pre>

Solved and explained in all details.


First equation of the given system represents an ellipse.

Second equation of the given system represents a hyperbola, &nbsp;having two separate branches.

Four solutions represent four intersection points of these figures.


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If you want to see many other similar &nbsp;(and different) &nbsp;problems solved, &nbsp;look into the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Systems-of-equations/Solving-the-system-of-algebraic-equations-of-degree-2.lesson>Solving systems of algebraic equations of degree 2</A> 

in this site.



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