Assume the equation has a solution for \( x \).

[tex]\[
\begin{array}{l}
n \cdot (17 + x) = 34x - r \\
x = \square + \bar{x}
\end{array}
\][/tex]



Answer :

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.