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When you encounter an equation with parentheses, such as \(-2(z-4)-(3z-2)\), it's a perfect opportunity to use the distributive property. The distributive property is a fundamental algebraic property that simplifies expressions and solves equations. Here’s how it works: multiply each term inside the parentheses by the term outside.
For example, in \(-2(z-4)\), both \(z\) and \(-4\) get multiplied by \(-2\). This results in \(-2z + 8\).
Similarly, \(-(3z-2)\) becomes \(-3z + 2\) when the distributive property is applied.
Once the parentheses are removed, it becomes much easier to further simplify and solve the equation. This step is crucial as it lays the foundation for combining like terms.

After applying the distributive property, the equation becomes \(-5z + 10 = -6z\). The next step is to solve for \(z\) by isolating it on one side of the equation.
Start by moving all terms involving \(z\) to one side and constants to the other.
1. Add \(6z\) to both sides to get rid of \(-6z\) on the right. This simplifies the equation to \(z + 10 = 0\).2. Subtract \(10\) from both sides to isolate \(z\). You now find \(z = -10\).
This systematic approach ensures that you don't lose track of any terms, and it showcases how step-by-step simplification helps solve complex equations. Remember, taking methodical steps keeps your work organized and increases accuracy.

After finding \(z = -10\), it’s crucial to check whether this solution is correct. Substitute \(z\) back into the original equation: \(-2(z-4)-(3z-2)=-2-(6z-2)\).
Simple substitution means replacing \(z\) with \(-10\). It helps verify if both sides of the original equation remain equal, confirming the solution’s validity.
1. Substitute and simplify the left side: \(-2(-10-4)-(3(-10)-2)\).
2. Simplify to see if it equals the right side after substitution.
Once both sides equal out, you ensure that \(z = -10\) is indeed the correct solution.
Checking solutions is a vital step in solving equations. It guarantees accuracy and reinforces the understanding of manipulation and substitution in algebra.