I find the given explanation obscure.
Start by naming things.
[MATH]a = \text {rate on machine A (in envelopes per hour}.[/MATH]
[MATH]b = \text {rate on machine B (in envelopes per hour}.[/MATH]
[MATH]c = \text {rate on machine C (in thousands of envelopes per hour}.[/MATH]
[MATH]h = \text {hours required for machine B to produce } 12,000 \text { envelopes.}[/MATH]
Any problem so far?
Four unknowns so we need four independent equations. We know that if rate is thousands of envelopes / hour, then rate times hours = thousands of envelopes.
[MATH]a = \dfrac{6}{3} = 2.[/MATH] Obvious.
[MATH]\dfrac{12}{5} * (b + c) = 6.[/MATH] Do you see where this came from?
[MATH]1 * (a + c) = 3.[/MATH] Should be obvious.
[MATH]h * b = 12.[/MATH] Obvious.
First solve for c.
[MATH]a = 2 \text { and } 1 * (a + c) = 3 \implies 2 + c = 3 \implies c = 1.[/MATH]
Now solve for b.
[MATH]\dfrac{12}{5} * (b + c) = 6 \text { and } c = 1 \implies \dfrac{12}{5} * (b + 1) = 6 \implies[/MATH]
[MATH]12(b + 1) = 30 \implies 12b = 18 \implies b = \dfrac{18}{12} = \dfrac{3}{2}.[/MATH]
And finally solve for h.
[MATH]h * b = 12 \text { and } b = \dfrac{3}{2} \implies \dfrac{3h}{2} = 12 \implies 3h = 24 \implies h = 8.[/MATH]