Question

Subtract 12(z+4)−3(14z+1) . Use fractions in final form.(1 point)
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Answer 1

To simplify the expression, we must distribute the coefficients to the terms inside the parentheses and combine like terms.

Starting with the first part of the expression, $12(z+4)$:
$$12(z+4)=12z+12(4)=12z+48$$

Now, let's simplify the second part, $3(14z+1)$:
$$3(14z+1)=3\cdot14z+3\cdot1=42z+3$$

Putting it all together, the expression becomes:
$$12(z+4)-3(14z+1)=(12z+48)-(42z+3)$$

Now, we can distribute the negative sign to the terms inside the second parentheses:
$$=12z+48-42z-3$$

Next, we can combine like terms by grouping the $z$ terms and the constant terms:
$$=(12z-42z)+(48-3)$$
$$=(-30z)+(45)$$

Thus, the final simplified form of the expression is $$-30z+45$$