Given $x=4a+3 \text{ and } x=7b+6,\;\; x,a,b \in \mathbb N,\;\; x,a,b > 0,\;$ find the minimum value for $x$.
How can I solve this system, given three unknown variables but only two equations?
Given $x=4a+3 \text{ and } x=7b+6,\;\; x,a,b \in \mathbb N,\;\; x,a,b > 0,\;$ find the minimum value for $x$.
How can I solve this system, given three unknown variables but only two equations?
If $a$ and $b$ are arbitrary real numbers, there's no such minimum since $4a+\frac{3}{4}$ and $7b+\frac{6}{7}$ both span the whole real line. Are you sure there are no additional restrictions on $a$ and $b$? For instance, if $a,b>0$, then the required minimum $x=4a+\dfrac{3}{4}=7b+\dfrac{6}{7}$ is $x=\frac{6}{7}$.
HInt:
$x = -1 \mod 4$ and $x = -1 \mod 7$.
If $a$ and $b$ are not related in any way, then you have an indeterminate system which allows infinite solutions for $x$.
EDIT: Since $a\in \mathbb{Z}$, if we consider $a$ to be non-negative, then the minimum possible value of $x$ would be with $a=1$, so $x=4+\frac{3}{4}=\frac{19}{4}$.
EDIT 2: New conditions have been added, $x,a,b\in \mathbb{N}$, therefore there's no answer, since the sum of a natural number ($4a$) plus a non-natural number ($\frac{3}{4}$) is always a non-natural number, and so therefore $x\not \in \mathbb{N}$ against our hypotheses.
EDIT 3: Problem has changed again. $4a+3 \equiv 6 \pmod 7$, then $4a \equiv 3 \pmod 7$, then the smallest $a$ that verifies this is $6$, so $x=27$, $b=3$.
Chinese Remainder Theorem .....
If both $x=4a+\frac{3}{4}$ and $x=7b+\frac{6}{7}$, then you have
$ 4a+\frac{3}{4}=7b+\frac{6}{7}. $
Then $a$ will be a function of $b$ or vice-versa. As fmartin says, this will admit an infinite number of solutions.