x + y = 9, x^2 - 3xy + 2y^2 = 0

Question

simultaneous non - linear equation... little confusing 2 me

Answer 1

x + y = 9 => y = 9-x

x^2 - 3xy + 2y^2 = 0
(x - y)(x - 2y)=0
x - y = 0 or x - 2y = 0
[This is a degenerate conic; i.e., two lines.]

For x-y=0:

x + y = 9
x - y = 0
2x = 9
x = 4.5
y = 4.5 b/c y = x

For x-2y=0:

2x + 2y = 18
 x - 2y = 0
3x = 18
x = 6
y = 9 - x = 3

Solutions are: (4.5,4.5) and (6,3)

check:

x + y = 9?         √               √

x^2 - 3xy    4.5^2(1-3+2)
+ 2y^2 = 0?      √        36-54+18 √

 

See a GeoGebra sketch here:

http://www.wyzant.com/resources/files/260649/degenerate_conic_line

Answer 2

Hi again Rodela;

x + y = 9, x^2 - 3xy + 2y^2 = 0

Let's isolate one variable in the first equation.  I randomly select x...

x=9-y

Let's take the second equation and substitute x with y-9...

x^2 - 3xy + 2y^2 = 0

(9-y)²-[(3y)(9-y)]+2y²=0

Let's FOIL (9-y)²...(9-y)(9-y)

FIRST...(9)(9)=81

OUTER...(9)(-y)=-9y

INNER...(-y)(9)=-9y

LAST...(-y)(-y)=y²

y²-9y-9y+81

y²-18y+81

y²-18y+81-27y+3y²+2y²=0

6y²-45y+81=0

Let's factor-out the 3...

3(2y²-15y+27)=0

The parenthetical equation must equal zero.  Therefore, we no longer need the 3...

2y²-15y+27=0

For the FOIL...

FIRST must be (2y)(y)=2y²

OUTER and INNER must add-up to -15y

LAST must be (9)(3) or (3)(9) or (1)(27) or (27)(1) and both numbers must be negative to render the sum of -15 and product of +27

(2y-9)(y-3)=0

Let's FOIL..

FIRST...(2y)(y)=2y²

OUTER...(2y)(-3)=-6y

INNER...(-9)(y)=-9y

LAST...(-9)(-3)=27

2y²-6y-9y+27=0

2y²-15y+27=0

(2y-9)(y-3)=0

Either or both parenthetical equations must equal zero...

2y-9=0, 2y=9, y=9/2

y-3=0,y=3

 

Let's establish the values of x by returning to the original equations...

y=3

x + y = 9, x^2 - 3xy + 2y^2 = 0

x+3=9...x²-[(3)(x)(3)]+[(2)(3²)]=0

x=6.......x²-9x+18=0

.............(x-6)(x-3)=0

Either or both parenthetical equation(s) must equal zero...

x-6=0

x=6

We are going to disregard x-3=0 because it does not correspond to the results of the other equation.

(6,3)

 

y=9/2

x + y = 9.......x^2 - 3xy + 2y^2 = 0

x+(9/2)=9.....x²-[(3x)(9/2)]+[(2)(-9/2)²]=0

x+9/2=18/2.....x²-(27/2)x+[(2)(81/4)]=0

x=9/2.........x²-(27/2)x+(81/2)=0

Let's multiply both sides by 2...

2x²-27x+81=0

(2x-9)(x-9)=0

Either or both parenthetical equation(s) must equal zero...

2x-9=0, 2x=9, x=9/2

We are going to disregard x-9=0 because it does not correspond to the results of the other equation.

(9/2,9/2)