Matt C.

ICM 5th period

Blackwell

Systems of Second Degree Equations and Inequalities

Given:

x² + 2y² = 10

and

3x² - y² = 9,

Solve the system of equations.

To solve this problem we have to find all the possible solutions for the system and to do this we are going to use the elimination method:

x² + 2y² = 10

3x² - y² = 9

First, multiply the top or bottom equation by a number so that one set of the coefficients are the same but of opposite sign.  We are going to solve for the y-coordinates first.

-3 ( x² + 2y² = 10)

3x² - y² = 9

which gives us:

-3x² - 6y² = -30

3x² - y² = 9

Next, add the two equations together and put answer into standard form (ax² + bx + c).

-7y² - 21 = 0

Now, to find the roots (the solutions for the 'y' value of the system) we use the Quadratic Formula.

(-b +/- sqrt( b² - 4ac)) / 2a

(0 +/- sqrt( 0 - 4(1)(-3)) / 2(1)

(+/- sqrt(12)) / 2

y = +/- sqrt(3) or +/- 1.73

Ok, now that we have the two possible solutions for the y-coordinates, we must solve for the x-coordinates.  In order to solve for the x-coordinate we can do one of two things: either repeat the above steps eliminating "y" out of the system or fill in the possible values of "y" into one of the original equations.

x² + 2y² = 10

2 (3x² - y² = 9)

So,

x² + 2y² = 10

6x² - 2y = 18

------------------------------

x² - (38/7) = 0

Now, we use the Quadratic Formula again to get the roots.

(0 +/- sqrt(0 - 4(1)(-38/7)) / 2(1)

+/- sqrt(152/7) / 2

x = +/- 2.3

So, the possible solutions for the system of equations is:

(+/- 2.3, +/- 1.7)

If you want some extra help go to:

 www.alltel.net/~okrebs/page65.html

www.smcc.cc.ms.us/fweb/mattc/SummerEveningCollAlg/6.3SysSecEqu.htm

maths.newcastle.edu.au/~mmmjr/Week13.pdf

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***quote was found at www.josephsoninstitute.com***