+0  
 
0
22
3
avatar+1347 

 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 64 + xy$.

 Jan 1, 2024
 #1
avatar+14 
+1

OK so,

lets first rearrange the second equation: x^2-xy+y^2=64.

and we know that x+y=10.

we can rewrite the second equation like this

(x+y)^2-3xy=64

and since x+y=10,

100-3xy=64

which means that xy=12.

NOW

we solve x+y=10 and xy=12.

we can substitute x as 10-y

so we have

y(10-y)=12

10y-y^2=12

and we get the quadratic

y^2-10y+12=0.

this isnt easily factorable so i will use the quadratic formula.

we get that y=5+sqrt 13, 5-sqrt 13.

so that means 

(x,y)=(5+sqrt 13, 5-sqrt 13) and (5-sqrt 13, 5+sqrt 13).

:)

 Jan 2, 2024
 #2
avatar+129316 
+1

Very nice solution, ethanyessir !!!

 

cool cool cool

 Jan 2, 2024

2 Online Users

avatar