He's WRONG, AGAIN, so:  

Answer by ikleyn(52781)   (Show Source): You can put this solution on YOUR website!
 .

Please help me solve 3x + y=3

xy =-6

~~~~~~~~~~~~~~







            The setup in the post by @josgarithmetic is  INCORRECT,  so ignore his post.



            I came to bring a correct solution.







    3x + y = 3     (1)

    xy = -6        (2)



From equation (1), express y = 3 - 3x = 3*(1-x)  and substitute it into equation (2). You will get

    3x*(1-x) = -6.



Simplify it step by step, making equivalent transformations

    x*(1-x) = -2

    x*(x-1) = 2

    x^2 - x - 2 = 0



Factor left side

    (x-2)*(x+1) = 0   


The roots are  x= 2  and  x= -1.


To get y, consider these two cases separately.


    a)  If x = 2,  then  y = 3*(1-x) = 3*(1-2) = 3*(-1) = -3.

    b)  If x= -1,  then  y = 3*(1-x) = 3*(1-(-1)) = 3*2 = 6.


ANSWER.  There are two solutions:  a) x=2,  y= -3   and   b) x= -1, y= 6.




Solved.







   

RELATED QUESTIONS

 Dear tutor, 
May you help me solve this question

Find dy/dx for each of the... (answered by Alan3354)

Please help me solve this problem:
y=1/2x-3
y=1/3x+2

I tried this
   y=1/3x+2... (answered by stanbon)

Please help me solve adding and subtracting polynomial: {{{(x^3 -3x^2y + 4xy^2 +... (answered by ichudov)

please help me solve the following:
(xy-y)(xy-y)=
(3x-5)(5x+5)=
(ab+c)(ab+c)=... (answered by funmath)

Please help me solve... (answered by stanbon)

Please help me solve for x and y simultaneously {{  y-2x+1 and... (answered by Alan3354)

Solve the initial value problem y'+y/2x=xy^-3,y(1)=2.Please help... (answered by t0hierry)

Solve the systems
a)x^2+y^2=25
  3x+y=13 

b) xy=20
   y=2x-3

c)  2x^2-xy+y^2=11
 (answered by lynnlo)

Solve the systems
a)x^2+y^2=25
  3x+y=13 

b) xy=20
   y=2x-3

c)  2x^2-xy+y^2=11
 (answered by lynnlo)