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Solve the System of Equations
3x^2-5y^2= 22
2x^2+4y^2 = 22

Solve the System of Equations 3x^2-5y^2= 22 2x^2+4y^2 = 22-example-1

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Lets try one way to solve this system. First, we want to isolate one of the unkown, lets choose y, so we need to remove x from an equation. In order to remove x from an equation, thats what we are going to do :

Line 1:

(3x² - 5y²) * 2 = 22 * 2

6x² - 10y² = 44

We simply multiplied Line 1 by the number 2, this doesn't change anything on the nature of Line 1.

Line 2:

(2x² + 4y²)*3 = 22*3

6x² + 12y² = 66

We simply multiplied Line 2 by the number 3, this doesn't change anything on the nature of Line 2.

Solving your system of equations is exactly the same thing as solving this new system of equation :

6x² - 10y² = 44

6x² + 12y² = 66

It is exactly the same. But it is easier now, because we can remove x from the equation simply by subtracting line 2 from line 1 :

L1 - L2 :

(6x² - 10y²) - (6x² + 12y²) = 44 - 66

6x² - 10y² - 6x² - 12y² = - 22

- 22y² = - 22

y² = 1

y = + 1 or y = - 1

We know have the value of y, we just have to replace it in the equation to solve the entire system :)

3x² - 5y² = 22

since y² will always be equal to one, wether y = 1 or y = - 1, we can write :

3x² - 5 = 22

3x² = 22 + 5

3x² = 27

x² = 27/3

x² = 9

x = + 3 or x = - 3

So the solutions are the following couples in the form (x,y) :

S = { (3 ; 1) , (-3;1) , (3 ; - 1) , (-3 ; -1))

Good Luck

User Chris Pink
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