Ok... so, we know that both pairs are conjugate of each other,
so.. both are complex solutions, namely, both are a+bi type
what's "a", we dunno, what is "b", we dunno either
but for z1, we know that a = m-2n-1 and b = 5m-4n-6
for z2, a = 3m-n-6 and b=m-5n-3
for those two folks to be conjugate of each other, they'd be
(a+bi)(a-bi)
notice, the "a"s are the same on z1 as well as z2
the "b" are the same also BUT the z2 "b" is negative,
in order to be a conjugate, regardless of that, the "b"s are the same
so, whatever (m-2n-1) is, is the same as <span>(3m-n-6), since a = a
and whatever </span><span>(5m-4n-6) is, will be the same value, since b =b,
the "a" and "b" in each conjugate, is the same value, thus one can say
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
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
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and pretty sure you can find "m" from there
once you have both, substitute in z1 or z2, to get the values for "a" and "b"</span>