Difficult Simultaneous Equation (1 Viewer)

[frenzal_dude]

Hi

How do you solve this equations analyticaly without solving a 4th degree polynomial
sqrt(x)+y=11; x+sqrt(y)=7

Thanks for the help!

 

[hscishard]

Well you could always graph it
Lmao idk, i'll be very interested in the answer though.

 

[cyl123]

By inspection x=4, y=9

 

[frenzal_dude]

hey, that's the right answer, but how did you get it without trial and error?

 

[khorne]

by inspection...i.e by looking at it...

 

[frenzal_dude]

so by inspection is the only way to get the answer?

 

[mitchy_boy]

You could try and do it old school, by squaring everything rearange, make x subject sub into equation 2, etc. Might take a while.

 

[Iruka]

You do what I would call enlightened trial and error:

(1) Sketch both graphs on the same number plane to confirm that there is only one solution. (Also gives you some limitation on the size of x and y.)

(2) Realize that the RHS of both equations is a whole number, which implies that x and y are both perfect squares.

(3) Try possible combinations of x and y until you get the one that works. From part (1), we know that the one that work is the only solution.

 

[cyl123]

You do what I would call enlightened trial and error:

(1) Sketch both graphs on the same number plane to confirm that there is only one solution. (Also gives you some limitation on the size of x and y.)

(2) Realize that the RHS of both equations is a whole number, which implies that x and y are both perfect squares.

(3) Try possible combinations of x and y until you get the one that works. From part (1), we know that the one that work is the only solution.

+1... (2) describes the mindset I had when I was finding real solutions. The fact there are perfect squares mean that x and y should be relatively small square numbers less that 11... hence 9 and 4.

Although complex solutions for x and y may exist....