Question

Given:
\(4x^2 + y^2 = 40\)
\(xy = 6\)

Find \(2x + y\).

Answer 1

Is this is a start?
x = 6/y

Answer 2

(2x+y)^2 = (4x^2 + y^2)+2xy
2x+y = ?

Answer 3

Yeah, that's what exactly is given in the book. How do you derive \((2x + y)^2\)?

Answer 4

That will make it lengthy parth..

Answer 5

I mean how did you get that \((2x + y)^2\)?

Answer 6

thats the start point...
we have a^2, b^2 and ab terms

and asking to find a+b

Answer 7

but there are four solution for (x,y)

Answer 8

(2x+y)^2-4xy=40
\[2x+y=\sqrt{40+24}\]
\[2x+y=\sqrt{64}=8\]

Answer 9

\[(2x + y)^2 = 4x^2 + y^2 + 4xy \implies 64\]

\[(2x + y) = \pm 8\]

Answer 10

it should be \[\pm 8\]

Answer 11

Hmm I am still understanding it.

Answer 12

\( \color{Black}{\Rightarrow (2x + y)^2 = 4x^2 +4xy + y^2 }\)

Answer 13

Parth,
first equation gives you : \(4x^2 + y^2 = 40\) and second \(xy = 6\)

Just plug these in the formula:
\[(2x + y)^2 = 4x^2 + y^2 + 4xy\]

Answer 14

Oh wait. Getting it.

Answer 15

\[(2x + y)^2 = \color{green}{(4x^2 + y^2)} + 4 \color{green}{(xy)} \implies (40) + 4(6)\]

Answer 16

\( \color{Black}{\Rightarrow (2x + y)^2 = 64}\)

\( \color{Black}{\Rightarrow 2x + y = \pm 8}\)

Answer 17

Yep

Answer 18

Yes now you got it right..

Answer 19

Thank you!

Answer 20

but we should get 4 values.. as per the pix of @UnkleRhaukus

Answer 21

Welcome dear..

Answer 22

scratch that... me confused :\