Question

It is given that (x+y)^2=16, (y+z)^2=36 and (z+x)^2=81 If x+y+z>3, then the number of possible value for x+y+z is

Answer 1

(x+y)^2=16 => x+y = +/- 4 - (a)
(y+z)^2=36 => y+z = +/- 6 - (b)
(z+x)^2=81 => z+x = +/- 9 - (c)

(a)+(b): x+2y+z = +/-4 +/-6 - (d)
(a)+(c): 2x+y+z = +/-4 +/-9 - (e)
(b)+(c): x+y+2z = +/-6 +/-9 - (f)

(d)+(e)+(f): 3x+3y+3z = 2(+/-4 +/-6 +/-9)
therefore: x + y + z = 2(+/-4 +/-6 +/-9)/3

that should help you arrive at the answer.

Answer 2

sorry - I think I made a mistake there, you can just add (a)+(b)+(c) to get:
2x + 2y + 2z = +/-4 +/-6 +/- 9
x+y+z = (+/-4 +/-6 +/-9)/2

Answer 3

then throw away all results that do not satisfy: x+y+z > 3

Answer 4

i don't understand, the way u write it down :(

Answer 5

\[x+y+z=\frac{\pm4\pm6\pm9}{2}\]

Answer 6

does that help?

Answer 7

A Mathematica solution is attached.

Answer 8

@pinky_cute1995 - which part is it that you do not understand?

Answer 9

the last part :
(d)+(e)+(f): 3x+3y+3z = 2(+/-4 +/-6 +/-9)
therefore: x + y + z = 2(+/-4 +/-6 +/-9)/3

Answer 10

ignore that - I made a mistake there.
If you just add equations (a), (b) and (c) then you get:\[2x + 2y + 2z = \pm4 \pm6 \pm9\]therefore:\[2(x+y+z)=\pm4 \pm6 \pm9\]

Answer 11

do you understand this step:\[(x+y)^2=16\implies x+y=\pm4\]

Answer 12

yes

Answer 13

ok, so if we do this for all three of your equations we end up with:\[(x+y)^2=16\implies x+y=\pm4\tag{a}\]\[(y+z)^2=36\implies y+z=\pm6\tag{b}\]\[(z+x)^2=81\implies z+x=\pm9\tag{c}\]understand so far?

Answer 14

yah

Answer 15

ok, so now we just add all equations (a)+(b)+(c) to get:\[x+y+y+z+z+x=\pm4\pm6\pm9\]therefore:\[2x+2y+2z=\pm4\pm6\pm9\]therefore:\[2(x+y+z)=\pm4\pm6\pm9\]make sense?

Answer 16

oh

Answer 17

so now make use of the fact you are given:\[x+y+z>3\]therefore:\[2(x+y+z)>6\]so see which of the combinations of:\[\pm4\pm6\pm9\]give you a value >6