If a+b+c = 76 and a-b-c = 38 , then what are the values of a, b, c ?
on solving both equation, 2a = 114 a = 57 but how 2 find b and c value
put the value of a in both the equations....you will get two equations in two variables b and c ....u can solve them
i get b+c=19 b+c=19 when i put a = 57 in both equations
when u put a=57 in second equation,you get b+c=-19
no, i get -b-c=-19 that is b+c=19
57-b-c=38 19=b+c ya right .....i am so sorry
it's k
in that case there is infinitely many solutions
b and c can have any values
and a=57
they came zero, when i solve them by 2312 method
but that method of cross multiplication is not applicable here
You've got two equations and three unknowns — not enough information to determine unique solution.
so vat would be b and c value.
b and c are natural numbers, do we have to assume that ?
no
Surely infinite solutions then.
k
essentially you have two planes that intersect at a common line. your solution will be the equation of that line.
the line itself has an infinite number of points on it :)
when a=0: b+c = 76 b-c = 38 --------- 2b = 114; b=57 ; c=19 gives us a point (0,57,19) when b=0: a+c = 76 a-c = 38 --------- 2a = 114; a=57 ; c=19 gives us a point (57,0,19) define the vector between the points and attach it to one of them to define the line of intersection. another method would be to cross the normals to define the vector and attach it to one of the points found.
errp, forgot to include the - on the b in the first parts :) might need to work that out better
hmmm..............then it would be 2b+2c = 114
b+c=57
it means the valus for b and c are infinite as @shubhamsrg has already said
For the values of b and c , there are infinite values, if we put b=1, c = 18 b=2, c=17 and so on..
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