Is this possible to solve through algebra?

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$$150 \equiv 17 \mod x, \qquad 100 \equiv 5 \mod x $$

Solve the simultaneous equation? Is this even a simultaneous equation? How do I find the value of $x$ too? I was doing a question and came up with these equations...I know the basics of modular arithmetic but don't really know too difficult ones

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Since $150-17=133$, we have $$ 150\equiv17\pmod{x}\implies x\mid133 $$ Furthermore, since $100-5=95$, we have $$ 100\equiv5\pmod{x}\implies x\mid95 $$ The possibilities for $x$ can be derived from the fact that $133=7\cdot19$ and $95=5\cdot19$.

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Hint:

The given equations imply $150-17=7\cdot19$ and $100-5=5\cdot19$ to be multiples of $x$. The $\gcd$ is $19$.

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If $a,b,c,d$ are specific numbers and $x$ is unknown (all positive integers) then the equation $a\equiv b\pmod x$ and $c\equiv d\pmod x$ by definition means $x$ is a common divisor of the two numbers $a-b$ and $c-d$.

The solution need not be unique. So solving them is the same as finding the set of all common divisors. This is same as finding $g=\gcd(a-b,c-d)$ and then listing all divisors of $g$.

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By definition your system is equivalent to $\ x\mid 133,\ x\mid 95,\,\ $ i.e. $\,\ x\mid 133,195$.

The universal property of the gcd tells us how to combine this pair of divisibility statements into an equivalent single divisibility statement, namely

$$ x\mid a,b \iff x\mid \gcd(a,b)$$