Solvable Simultaneous Equations

Adding the first solution to twice of the second, we have $(a-4c)x = 3$ . From the third equation, we have $x = c- 25$ , which gives $(c-25)(a-4c) = 3$ . Since $a,c$ are integers this implies that $c-25$ and $a-4c$ both are integers and since $x$ is positive implies that $c-25$ is positive and so is $a-4c$ .

CASE 1: $c-25=1 , a-4c=3 \Rightarrow \ c = 26 , a=107$ .

CASE 2: $c-25=3 , a-4c=1 \Rightarrow \ c=28 , a= 113$

So we conclude that $\min(a)=107$ .

[Edits for clarity - Calvin]

Since $a, b$ and $c$ are positive integers and $x$ is positive, we must have $c \geq 26, (b-2c) \geq 1$ and $(a-2b) \geq 1$ .

This implies that $a \geq 2b + 1\geq 2(2c+1) + 1 \geq 2 ( 2 (26) + 1 ) + 1 = 107$ .

Note that $(a, b, c) = (107, 53, 26)$ has a solution at $x = 1$ which is positive. Hence the minimum value of $a$ is $107$ .

(a−2b)x=1/ (b−2c)x=1
a-2b=b-2c
a=3b-2c....(4)
from x+25=c from (3)
and b=1/x+2c from (2)
Substituting the value in
a=3b-2c
we get,
a=3/x+4x+100
for a to be minimum x should also be minimum
therefore,for x=1(least +ve soln)
a=3+4+100
a=107

The three equations are given such that x= \frac {1}{a-2b} = \frac {1}{b-2c} = c-25

As x is positive, the smallest value of a would also hold the smallest values of b and c from the relations described in the first two simulateous equations.

The third simultaneous equation can then be used to find the smallest possible value for the integer c, which is 26 such that x=1. Applying this into the second equation, the smallest value of b is 53, and the smallest value of a is thus 107.

First putting the value of x in (b-2c)x=1 (b-2(x+25))x=1 b-2x2-50x=1 b-1=2x2+50x b=2x2+50x+1 From here we will get the value of b in terms of x and then putting this value of b in (a-2b)x=1, we will get the value of a in terms of x (a-2(2x2+50x+1))x=1 a-4x3-100x2-2x=1 and as x is positive and the minimum value of x as positive will be 1 so putting 1 in place of x in the a a=1+4+100+2 a=107 we g0t the minimum value of a.

we know that $\large x+25=c$ and x is positive number so x minimum make a is minimum so x=1

and we can find c=26

$\large (a-2b)x=1$ $\large (a-2b)x=(b-2c)x$ $\large x\neq0$ $\large a+2c=3b$

$\large (b-2c)x=1$ x=1 $\large (b-2c)=1$ $\large b=2c+1$ $\large b=53$

$\large a+2c=3b$ $\large a+2.26=3.53$ so $\large a=107$