Nilesh S.

asked • 09/15/16

Miscellaneous Equations

(x+y)^2 + 3(x-y)=30; xy + 3(x-y)=11 what are the values of x and y ?

Don L.

tutor
Hi Nilesh, I believe the first term given is incorrect: (x=y)2. Should this be (x+y)2?
 
Let us know.
 
Don
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09/15/16

Nilesh S.

Yes you are right indeed it is (x+y)^2 + 3(x-y) = 30 ;
                                           xy + 3(x-y)=11
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09/15/16

Jason L.

I will admit I just spent 30 minutes working on this before I just threw my hands up.  The answer is going to be very complicated.  I'll leave my work here just in case someone wants to keep going (or figure out a place I went wrong):
 
(x+y)^2 + 3(x-y)=30
xy + 3(x-y)=11

(x^2 + 2xy + y^2) + (3x - 3y) =30
xy + (3x - 3y)=11

Your goal is to isolate a variable to one side. Using eq2...

3x + xy = 11 + 3y
x(3+y) = 11 + 3y
x = (11 + 3y)/(3+y)

Now plug into eq1...

[(11+3y)/(3+y)]^2 + 2y[(11 + 3y)/(3+y)] + y^2 + 3[(11 + 3y)/(3+y)] - 3y =30
(11+3y)^2/(3+y)^2 + 2y(11+3y)/(3+y) + 3(11+3y)/(3+y) = -y^2 + 3y + 30

**multiply both sides by (3+y)/(11+3y)**
(11+3y)/(3+y) + 2y + 3 = (-y^2 + 3y + 30)*(3+y)/(11+3y)

**multiply left by (11+3y) and right by (3+y) to get rid of denominators**
(11+3y)^2 + 2y(11+3y) + 3(11+3y) = (-y^2 + 3y + 30)*(3+y)^2

(11+3y)[11+3y + 2y + 3] = (-y^2 + 3y + 30)*(3+y)^2
(11+3y)[5y + 14] = (-y^2 + 3y + 30)*(3+y)^2
97y + 154 + 15y^2 = (-y^2 + 3y + 30)*(3+y)^2
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09/15/16

1 Expert Answer

By:

Ronay R. answered • 09/15/16

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Nilesh, you replied to Don's comment as "Yes you are right indeed it is (x+y)^2 + 3(x-y) = 30 ;
xy + 3(x-y)=11", but these are the same equations you gave at the very beginning. So, the first equation is correct or not? According to your reply, I am going to work on it.
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