1Cademy - Solving $$12[1 - 5(4z - 1)] = 3(24 + 11z)$$ with Nested Grouping Symbols

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Strategy for Solving Linear Equations

Example

Solving $12[1 - 5(4z - 1)] = 3(24 + 11z)$ with Nested Grouping Symbols

To solve the algebraic equation $12[1 - 5(4z - 1)] = 3(24 + 11z)$ , we begin by simplifying the expression starting from the innermost parentheses. Distributing $-5$ into $(4z - 1)$ updates the equation to $12[1 - 20z + 5] = 3(24 + 11z)$ . Combining the constant integers inside the brackets yields $12[6 - 20z] = 3(24 + 11z)$ . We then distribute the outer coefficients across both binomial expressions to get $72 - 240z = 72 + 33z$ . Adding $240z$ to both sides collects the variable terms on the right, yielding $72 = 72 + 273z$ . Subtracting $72$ from both sides groups the constants on the left, resulting in $0 = 273z$ . Ultimately, dividing both sides by $273$ yields the final valid solution $z = 0$ . This is an additional problem corresponding to 'Try It 2.10', serving as hands-on practice for systematically solving linear equations.

Updated 2026-05-02

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OpenStax Intermediate Algebra

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