Example

Solving 6[42(7y1)]=8(138y)6[4 - 2(7y - 1)] = 8(13 - 8y) with Nested Grouping Symbols

To solve the equation 6[42(7y1)]=8(138y)6[4 - 2(7y - 1)] = 8(13 - 8y), we start by simplifying the innermost parentheses. Distributing 2-2 through (7y1)(7y - 1) gives 6[414y+2]=8(138y)6[4 - 14y + 2] = 8(13 - 8y). Combining the constant terms inside the brackets results in 6[614y]=8(138y)6[6 - 14y] = 8(13 - 8y). Distributing the factors on both sides yields 3684y=10464y36 - 84y = 104 - 64y. By adding 84y84y to both sides, we gather the variable terms on the right side: 36=104+20y36 = 104 + 20y. Subtracting 104104 from both sides isolates the constant terms on the left as 68=20y-68 = 20y. Dividing both sides by 2020 finishes isolating the variable, resulting in y=6820y = -\frac{68}{20}, which simplifies to the final solution y=175y = -\frac{17}{5}. This provides a practice example on applying a general strategy to solve linear equations involving nested grouping symbols.

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Updated 2026-05-02

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