Sure! Let's solve the given equation step by step and assume it has a solution for \( x \). The equation is:
[tex]\[
n \cdot (17 + x) = 34x - r
\][/tex]
where \( x = \square + \bar{x} \).
Step 1: Expand and rearrange the equation:
[tex]\[
n \cdot 17 + n \cdot x = 34x - r
\][/tex]
Step 2: Collect the \( x \) terms on one side of the equation:
[tex]\[
n \cdot 17 = 34x - r - n \cdot x
\][/tex]
Step 3: Factor out the \( x \) terms on the right-hand side:
[tex]\[
n \cdot 17 = x(34 - n) - r
\][/tex]
Step 4: Isolate \( x \):
[tex]\[
x(34 - n) = n \cdot 17 + r
\][/tex]
[tex]\[
x = \frac{n \cdot 17 + r}{34 - n}
\][/tex]
Step 5: Decompose \( x \) into its integer part (\( \square \)) and its fractional part (\( \bar{x} \)):
Let \( x = k + \bar{x} \), where \( k \) is an integer and \( 0 \leq \bar{x} < 1 \).
Thus,
[tex]\[
k = \left\lfloor \frac{n \cdot 17 + r}{34 - n} \right\rfloor
\][/tex]
and
[tex]\[
\bar{x} = \frac{n \cdot 17 + r}{34 - n} - k
\][/tex]
In summary, the equation:
[tex]\[
n \cdot(17 + x)=34 x-r
\][/tex]
has its solution for \( x \) expressed as:
[tex]\[
x = \frac{n \cdot 17 + r}{34 - n}
\][/tex]
where \( x = k + \bar{x} \):
- \( k = \left\lfloor \frac{n \cdot 17 + r}{34 - n} \right\rfloor \)
- \( \bar{x} = \frac{n \cdot 17 + r}{34 - n} - k \)
This completes the step-by-step solution for decomposing [tex]\( x \)[/tex] into its integer part and its fractional part.
