question
Solve the equation $\frac{38-11x}{6}=\{\frac{2x+1}{6}\}+\{\frac{x+17}{3}\}$ in the set of real numbers.
my idea
Because ${a}$ is between $1$ and $0$ we can say that $0\leq\frac{38-11x}{6}<2$
after some processing, we can say that
$\frac{38}{11}\geq x>\frac{26}{11}$
Also we know that $x=[x]+\{x\}$ so we can write the equality as
$[\frac{2x+1}{6}]+[\frac{x+17}{3}]=\frac{15x-3}{6}\in Z$
which means that $15x-3=6k => 5x=2k+1$, which basically means that $x$ is even.
(*) Using @Robert Israel's idea we can write that
$[\frac{2x+1}{6}]$ can be $0$ or $1$ and $[\frac{x+17}{3}]$ can be only $6$.
Taking each case we get that $x=\frac{13}{5}$ or $3$ but the thing is that $x=\frac{13}{5}$ isn't working if we substitute x in the main equality.
Can you please tell me if my idea is ok and where I was wrong?
I hope one of you can help me! Thank you!
Best Answer
$$\frac{\mathrm{38}−\mathrm{11}{x}}{\mathrm{6}}=\left\{\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{6}}\right\}+\left\{\frac{{x}+\mathrm{17}}{\mathrm{3}}\right\} \\ $$ $$\left[\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{6}}\right]+\left[\frac{{x}+\mathrm{17}}{\mathrm{3}}\right]=\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{2}}\in{N} \\ $$ $$\frac{\mathrm{38}−\mathrm{11}{x}}{\mathrm{6}}=\left[\mathrm{0};\mathrm{2}\right) \\ $$ $$\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{2}}=\left(\frac{\mathrm{119}}{\mathrm{22}};\frac{\mathrm{179}}{\mathrm{22}}\right] \\ $$ $$\frac{\mathrm{5}{x}−\mathrm{1}}{\mathrm{2}}=\left\{\mathrm{6},\mathrm{7},\mathrm{8}\right\} \\ $$ $${x}=\left\{\cancel{\frac{\mathrm{13}}{\mathrm{5}}},\mathrm{3},\cancel{\frac{\mathrm{17}}{\mathrm{5}}}\right\} \\ $$