To solve the equation \( 6(3z - 2) + 4(z - 5) = 78 \), we'll start by distributing the terms on the left side.
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Distribute \(6\) in the first term: \[ 6(3z) - 6(2) = 18z - 12 \]
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Distribute \(4\) in the second term: \[ 4(z) - 4(5) = 4z - 20 \]
Now, the equation becomes: \[ 18z - 12 + 4z - 20 = 78 \]
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Combine like terms (\(18z\) and \(4z\), \(-12\) and \(-20\)): \[ (18z + 4z) + (-12 - 20) = 78 \] \[ 22z - 32 = 78 \]
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Add \(32\) to both sides to isolate the term with \(z\): \[ 22z - 32 + 32 = 78 + 32 \] \[ 22z = 110 \]
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Divide both sides by \(22\): \[ z = \frac{110}{22} = 5 \]
Therefore, the solution to the equation is: \[ \boxed{5} \]