Question

solve

Answer 1

\[\sqrt{2x-1}+\sqrt{y+3}=3\]
\[2xy+6x-y=7\]

Answer 2

seems like (2)\[2x(y+3)-y=7\]

Answer 3

\[p=\sqrt{y+3}\]
\[q=\sqrt{2x-1}\]

Answer 4

\[p+q=3\]
\[(q^2+1)p=7+y\]

Answer 5

no 2nd 1\[p^2(q^2+1)=7+y\]

Answer 6

\[(pq)^2+p=7+p^2-3\]

Answer 7

\[(pq)^2-p^2+p-4=0\]
\[p+q=3\]

Answer 8

hey once slow down and think,

Answer 9

ok

Answer 10

can sum of 2 positive irrational numbers give you a rational number ?

Answer 11

no

Answer 12

stickly speaking it can.. you think about it, in this case no it can not, so what are possible combinations fot 3..

Answer 13

@hartnn

Answer 14

\[p^2(p-3)^2-p^2+p-4=0\]
\[p^4-6p^3+8p^2+p-4=0\]

Answer 15

1+2, 2+1, 3+0,0+3 right, so out of possible (x,y) pairs are (1,1) satisfies both equations write (i am assuming you know tha traditional way to solve)

Answer 16

we should have eliminated the y also in the beginning

Answer 17

\[p=1\]
\[1=y+3,y=-2\]
\[2=q,x=5/2\]

Answer 18

seconf equation we may rewrite as
\(2xy + 6x - y = 7\)
\(y(2x-1) + 6x = 7\)
\(y(2x-1) + 3(2x-1) = 7-3\)
\((2x-1)(y+3) = 4\)

Answer 19

this is nice

Answer 20

then subbint p,q we get :

\(p+q = 3\)

\(p^2q^2 = 4\)

Answer 21

but did you noticethat i eliminated y\[y=p^2-3\]

Answer 22

ha yes... i see nothing wrong in that method also

Answer 23

but it made the equation clumsy

Answer 24

\[p+q=3,pq=\pm 2\]

Answer 25

p >= 0, q >= 0
so we -2 we can ignore

Answer 26

\[p^2-3p+2=0\]
\[q^2-3q+2=0\]

Answer 27

\[p=q=1,p=q=2,\]

Answer 28

tired now i'll slve later

Answer 29

dude it is p=1 and q=2