For what values of b does the following system of modular equations have a solution?

Question

$$x \equiv b (100)$$ $$x \equiv b^2 (35)$$ $$x \equiv 3b - 2 (49)$$

If I was pressed for an answer I would say this system was unsolvable. If $x \equiv b(100)$ the $x = b + 100t$. Then $b + 100t \equiv b^2 (35)$ which would imply that $100t \equiv b^2-b(35)$. The problem here is 100 is an invertible element in mod(35). Is this line of thought correct?

Answer 1

HINT:

$$x\equiv3b-2\pmod{49}\implies x\equiv3b-2\pmod7$$

Again, $$x\equiv b^2\pmod{35}\implies x\equiv b^2\pmod7$$

$\displaystyle\implies b^2\equiv3b-2\pmod7\iff(b-1)(b-2)\equiv0\pmod7$

As $(b-1,b-2)=1$ either $7|(b-1)\iff b\equiv1\pmod7,b=7t+1$ where $t$ is any integer

or $b\equiv2\pmod7$

Answer 2

At least for $b=1$ you have the solution $x=1$.