Solve for z-6z-1- 3z-2

Given 6z+1 ≤ 3(z 2) Which implies 6z+1≤3z 6 , which gives 3z ≤ 7 Therefore z ≤ 7/3 ...

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Modulus of a Complex Number

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Question

Solve for $z : 6 z + 1 \leq 3 (z - 2)$

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Solution

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z .

\frac{4}{z - 1} - \frac{5}{z + 2} = \frac{3}{z}, z \neq 1, 0, - 2

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5 x - 6 y + 4 z = 157 x + 4 y - 3 z = 192 x + y + 6 z = 46

$S o l v e : 4 y z (z^{2} + 6 z - 16) \div 2 y (z + 8)$

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