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(C) Let $u = |y|$ and $v = [z]$. The system becomes: $2x + 3u + 5v = 0$ $x + u - 2v = 4$ $x + u + v = 1$ Subtracting the second equation from the thir

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infinitely many solutions

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(C) Let $u = |y|$ and $v = [z]$. The system becomes: $2x + 3u + 5v = 0$ $x + u - 2v = 4$ $x + u + v = 1$ Subtracting the second equation from the third: $(x + u + v) - (x + u - 2v) = 1 - 4$ $3v = -3 \Rightarrow v = -1$. Substituting $v = -1$ into the second and third equations: $x + u + 2 = 4 \Rightarrow x + u = 2$ $x + u - 1 = 1 \Rightarrow x + u = 2$. Since both equations reduce to $x + u = 2$,we have infinitely many pairs of $(x, u)$ satisfying this. Given $u = |y| = 2 - x$,for any $x \leq 2$,$u$ is non-negative. Also,$v = [z] = -1$ implies $z \in [-1, 0)$. Since $x$ can take any value such that $|y| = 2 - x \geq 0$ (i.e.,$x \leq 2$),there are infinitely many solutions for $(x, y, z)$.

If $[x]$ is the greatest integer less than or equal to $x$ and $|x|$ is the modulus of $x$,then the system of three equations $\begin{aligned} & 2x + 3|y| + 5[z] = 0, \\ & x + |y| - 2[z] = 4, \\ & x + |y| + [z] = 1 \end{aligned}$ has

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