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Ta-3(a-1) = 3+a %3D

Ta-3(a-1) = 3+a %3D

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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The image contains the following algebraic equation written on lined paper: \[ 12 - 3(a - 1) = 3 + a \] This is a linear equation which involves a single variable, \( a \). The goal is typically to solve for \( a \). ### Explanation: 1. **Equation Breakdown**: - **Left Side**: \( 12 - 3(a - 1) \) - Here, \( 3 \) is being multiplied by the expression \( (a - 1) \). - **Right Side**: \( 3 + a \) - This side adds 3 to the variable \( a \). 2. **Solving the Equation**: - First, distribute the \(-3\) across the terms inside the parentheses on the left side: - \( 12 - 3 \times a + 3 = 3 + a \). - Simplify to \( 12 - 3a + 3 = 3 + a \). - Combine like terms: - \( 15 - 3a = 3 + a \). - Next, isolate \( a \) by adding \( 3a \) to both sides: - \( 15 = 3 + 4a \). - Subtract 3 from both sides to isolate terms with \( a \): - \( 12 = 4a \). - Finally, divide both sides by 4 to solve for \( a \): - \( a = 3 \). The solution to the equation is \( a = 3 \).
Transcribed Image Text:The image contains the following algebraic equation written on lined paper: \[ 12 - 3(a - 1) = 3 + a \] This is a linear equation which involves a single variable, \( a \). The goal is typically to solve for \( a \). ### Explanation: 1. **Equation Breakdown**: - **Left Side**: \( 12 - 3(a - 1) \) - Here, \( 3 \) is being multiplied by the expression \( (a - 1) \). - **Right Side**: \( 3 + a \) - This side adds 3 to the variable \( a \). 2. **Solving the Equation**: - First, distribute the \(-3\) across the terms inside the parentheses on the left side: - \( 12 - 3 \times a + 3 = 3 + a \). - Simplify to \( 12 - 3a + 3 = 3 + a \). - Combine like terms: - \( 15 - 3a = 3 + a \). - Next, isolate \( a \) by adding \( 3a \) to both sides: - \( 15 = 3 + 4a \). - Subtract 3 from both sides to isolate terms with \( a \): - \( 12 = 4a \). - Finally, divide both sides by 4 to solve for \( a \): - \( a = 3 \). The solution to the equation is \( a = 3 \).
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