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How to find X from2y^2+6y=9x. And y^2-2y=3x?

Author :

Submitted : 2018-06-14 08:00:32    Popularity:     

Tags: find  from2y  

2y^2 + 6y = 9x y^2 - 2y = 3x 3y^2 - 6y = 2y^2 + 6y y^2 - 12y = 0 y(y - 12) = 0 If y = 0, x = 0 If y = 12, x = 40 Make any variable the sub

Answers:

2y^2 + 6y = 9x
y^2 - 2y = 3x
3y^2 - 6y = 2y^2 + 6y
y^2 - 12y = 0
y(y - 12) = 0
If y = 0, x = 0
If y = 12, x = 40

Make any variable the subject of the formula and substitute to the other

2y^2 + 6y = 9x.
y^2 - 2y = 3x
2y^2 - 4y = 6x
10y = 3x
y^2 - 2y = 10y
y(y - 12) = 0
Solutions:
y = 0
y = 12
If y = 0, x = 0
If y = 12, x = 40

2y^2+6y=9x
x = (2y^2+6y) /9

y^2-2y=3x
x = (y^2-2y)/3
x = (3y^2-6y)/9

Equate the two x's

(2y^2+6y)/9 = (3y^2-6y)/9
2y^2+6y = 3y^2-6y
-y^2+12y =0
multiply both sides by -1
y^2-12y=0
y(y-12)=0
y=0
y=12

x= (2y^2+6y)/9
when y=0 :
x= ((2)(0)+(6)(0))/9 = 0

when y=12
x = ((2)(144)+(6)(12))/9
x = (288+72)/9
x= 360/9
x=40

x=0; x=40

3x = y^2 - 2y...[1].
9x =2y^2 + 6y...[2].
{[2]-3[1]}--> 0 = (2y^2 +6y)-(3y^2 -6y) = 12y-y^2 = y(12-y). y =/= 0. Otherwise system collapses. y = 12.
Then [1]--> x = (1/3)(12^2 - 24) = 40. End of story.

2y² + 6y = 9x ← this is a hyperbola

y² - 2y = 3x ← this is another hyperbola → you multiply by 3 both sides

3y² - 6y = 9x


9x = 9x

2y² + 6y = 3y² - 6y

2y² - 3y² = - 6y - 6y

y² - 12y = 0

y.(y - 12) = 0


First case: y = 0

Recall: y² - 2y = 3x

0 - 0 = 3x

x = 0

First solution = { 0 ; 0 }


Second case: (y - 12) = 0 → y - 12 = 0 → y = 12

Recall: y² - 2y = 3x

144 - 24 = 3x

3x = 120

x = 40

Second solution = { 40 ; 12 }

2y^2+6y = 3y^2-6y
y^2-12y = 0
y(y-12) = 0
y = 12 or 0
x = 40 or 0
Intersection points (40,12) & (0,0)
See Graph
https://www.desmos.com/calculator/5oquzl...

2y²+6y = 9x = 3[3x] = 3[y²-2y]
2y²+6y = 3y²-6y ----->12y = y² -----> 12 = y
[2*12²+6*12]/9 = x = 40



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