6(3z-2)+4(z-5) =78

1 answer

Let's solve the equation step by step:

Starting with the equation:

\[ 6(3z - 2) + 4(z - 5) = 78 \]

First, distribute the \(6\) and \(4\):

\[ 6 \cdot 3z - 6 \cdot 2 + 4 \cdot z - 4 \cdot 5 = 78 \] \[ 18z - 12 + 4z - 20 = 78 \]

Next, combine like terms:

\[ (18z + 4z) + (-12 - 20) = 78 \] \[ 22z - 32 = 78 \]

Now, add \(32\) to both sides:

\[ 22z - 32 + 32 = 78 + 32 \] \[ 22z = 110 \]

Next, divide both sides by \(22\):

\[ z = \frac{110}{22} \] \[ z = 5 \]

Thus, the solution for \(z\) is:

\[ \boxed{5} \]