How to solve ratios in maths?

Question

A varies directly as sum of two quantities b and c .b in turn varies directly as x and c varies inversely as x. when x=1 or 2,a=3. find the value of a when x=4

Answer 1

A varies directly as sum of two quantities b and c

A = k1(b+c)

b in turn varies directly as x

   b = k2 * x

c varies inversely as x

   c  = k3/x

sub val of b and c in eq 1

A  = k1(k2 x +  k3/x)   put k1k2  = c1  k1k3  = c2

so

A  = c1x+c2/x

x = 1  a = 3   implies  3 = c1+c2

x = 2  a = 3   implies  3  = 2c1 +  c2/2

solving we get  

            c1  = 1  and c2  = 2

a  = x+  2/x

a  = 4  +  2/4  = 9/2  = 4.5

Answer 2

... difficult to understand your problem, it's not very clear to me... here's what I've got, but I hope someone else helps you.

where Y and Z are constants.

A=B+C

B=YX

C=Z/X

Then A=YX+Z/X

so then system of equations by plugging in the values...

3=Y1+Z/1

3=Y2+Z/2

then solve by multiplying then subtraction and

A=X+2/X

so then where X=4

A=4+2/4

A=4.5