341Also a+b+c = \frac{10}{a} = \frac{12}{b} = \frac{14}{c} = \frac{10+12+14}{a+b+c} = \frac{36}{a+b+c}
Hence a+b+c = ±6
This is for class IX X only
a(a+b+c) = 10
b(a+b+c) = 12
c(a+b+c) = 14
Find a, b, c
-
UP 0 DOWN 0 0 13
13 Answers
1a/b = 5/6
b/c = 6/7
c/a = 7/5
so (a,b,c) = (5x , 6x , 7x)
5x(18x) = 10
so x= ±1/3
so (a,b,c) = (±5/3 , ± 2 , ±7/3 )
341
62I guess this is the first time prophet sir missed a simpler solution?
or is it that you were showing an alternative solution?
just adding the three equations directly gives us
\left(a+b+c \right)^2=36
Hence a+b+c=±6
hence the solution :)
341:D not the first time I have come up with a convoluted solution
62@fibonaci
(a,b,c) = (±5/3 , ± 2 , ±7/3 ) is not the right answer....
instead the righta answer shoudl be (a,b,c) = ±(5/3 ,2 ,7/3)
@Prophet sir: ;)
1that is what i meant
either all positive at a time or all negative at a time
110/a=12/b=14/c
b=6a/5; c=7a/5
putting this in the first equ.
we get,a(a+6a/5+7a/5)=10
solving,we get a=5/3
similiarly we will get b and c
36adding all the equations we get,
(a + b + c) = 6 or (a + b + c) = -6
Now, let us consider any of the three equations
firstly a(a + b + c) = 10
=> a + b + c = 10/a
=> b + c = (10 - a2)/a
=> 6 - a = (10 - a2)/a and -6 - a = (10 - a2)/a
=> 6a - a2 = 10 - a2 and -6a - a2 = 10 - a2
=> a = 5/3 or a = -5/3
and similiarly we get, a , b and c......
36I must truly say i m a gadha....!!
Simple solution....
On adding all the three eqns. we get,
a + b + c = 6 or, a + b + c = -6
On putting these values of a + b + c in
all the three eqns. we get the values of a,b and c... simple...!!
1hehe n i must truly agree u r one![3][3][3][3]...just kidding man...no issues!
36@kunl -> How do u expect me to feel bad over something which is "a universal truth...lol".??
