Finding $c-a$ from $2$ equations

Question

Given that $$2017(a-b)+2018(b-c)+2019(c-a)=0,$$ and that $$2017^2(a-b)+2018^2(b-c)+2019^2(c-a)=2018,$$ find $c-a.$

I was able to rearrange the first equation into $b+c-2a=0$, and that the second equation is $$4035b+4037c-8072a=2018.$$ I am stuck here. I also cannot directly find $c$ or $a$ since there are only $2$ equations.

Answer 1

From the first equation, we find:

$$c - a = a - b$$

We also find:

$$c = 2 a - b \iff b - c = b - 2 a + b = 2 b - 2 a = -2 (a - b)$$

Let $x = c - a$. Based on the second equation, we find:

$$2017^2 x - 2 \cdot 2018^2 x + 2019^2 x = 2018 \iff 2 x = 2018 \iff x = 1009$$

Answer 2

Calling $a-b = x, b-c = y , c-a = z$ we have the linear system

$$ 2017 x+ 2018 y + 2019 z = 0\\ 2017^2 x+ 2018^2 y + 2019^2 z = 2018\\ x + y + z = 0 $$

and solving for $x,y,z$ we have

$$ x=1009\\ y=-2018\\ z=1009 $$