Problem 39 - Algebra 2

Question

Solve each equation. Check your solutions. \(3|2 a+7|=3 a+12\)

Step-by-Step Solution

Answer

The solutions are \( a = -3 \) and \( a = -\frac{11}{3} \).

Step 1: Split the Absolute Value

Start by understanding that the equation involves an absolute value expression \( |2a + 7| \). This means we need to consider two cases: once where the expression inside the absolute value is positive and once where it is negative. This leads us to two separate equations. For the first case, assume \( 2a + 7 \geq 0 \) which simplifies to \( 2a + 7 = 2a + 7 \) and for the second case \( 2a + 7 < 0 \) which will be \( |2a + 7| = -(2a + 7) \).

Step 2: Solve for Case 1

For the first situation, the equation becomes \( 3(2a + 7) = 3a + 12 \). Distribute the 3 on the left to get \( 6a + 21 = 3a + 12 \). Solve for \( a \) by first subtracting \( 3a \) from both sides to get \( 3a + 21 = 12 \). Then, subtract 21 from both sides to get \( 3a = -9 \). Finally, divide by 3 to find \( a = -3 \).

Step 3: Solve for Case 2

For the second situation, the equation becomes \( 3(-(2a + 7)) = 3a + 12 \). This can be simplified to \( -6a - 21 = 3a + 12 \). To solve for \( a \), first add \( 6a \) to both sides to obtain \( -21 = 9a + 12 \). Subtract 12 from both sides to have \( -33 = 9a \). Finally, divide both sides by 9 to find \( a = -\frac{11}{3} \).

Step 4: Check the Solutions

Substitute \( a = -3 \) back into the original equation. The left side becomes \( 3|2(-3) + 7| = 3| -6 + 7| = 3|1| = 3 \). The right side is \( 3(-3) + 12 = -9 + 12 = 3 \). Hence, \( a = -3 \) is a valid solution. Now, substitute \( a = -\frac{11}{3} \) back into the original equation. The left side becomes \( 3|2(-\frac{11}{3}) + 7| = 3| -\frac{22}{3} + \frac{21}{3}| = 3| -\frac{1}{3}| = 1 \). The right side evaluates to \( 3(-\frac{11}{3}) + 12 = -11 + 12 = 1 \). Thus, \( a = -\frac{11}{3} \) is also a valid solution.

Step 5: Conclusion: Solution Set

After checking, both solutions solve the original equation, meaning the solution set includes both \( a = -3 \) and \( a = -\frac{11}{3} \).

Key Concepts

Solution Sets

In the context of absolute value equations, like the example given, the solution set refers to all values of the variable that satisfy the equation. A typical absolute value equation breaks into two distinct equations, owing to how absolute values can represent both positive and negative outcomes of an expression. Here, the equation \(3|2a + 7| = 3a + 12\) splits into two cases:

• For one part, assume the expression inside the absolute value is positive or zero: \(2a + 7 \geq 0\), leading to a straightforward equation.
• For the second part, assume that \(2a + 7 < 0\), meaning the equation changes by reflecting the negative condition: \(|2a + 7| = -(2a + 7)\).

The solution set then comprises all `\(a\)` values solved from both equations that satisfy the original equation. As seen, the solutions would be \(a = -3\) and \(a = -\frac{11}{3}\). This is an important result as sometimes one or both solutions don't work, hence the importance of our next concept: checking solutions.

Checking Solutions

Checking solutions is a crucial step to avoid extraneous solutions. When dealing with absolute value equations, you solve derived cases, but some solutions might not fit the original settings. Here's how to verify:

• Substitute each found solution back into the original equation.
• Calculate both sides of the equation to confirm they are equal.

By substituting \(a = -3\) and \(a = -\frac{11}{3}\) into \(3|2a + 7| = 3a + 12\):- For \(a = -3\), the left side becomes \(3|1| = 3\) and the right side becomes \(3 = 3\).- For \(a = -\frac{11}{3}\), both sides simplify to \(1\).Thus, both are correct, ensuring our solution set is valid, affirming the solutions indeed satisfy the given equation.

Solving Linear Equations

Solving linear equations is often the key part of breaking down absolute value problems. Each split equation from the absolute value forms a linear equation. These have the general form \(ax + b = c\), where you solve for the unknown variable:

• Isolate the variable on one side by performing inverse operations: addition, subtraction, multiplication, or division.
• Simplify progressively, reducing terms until the variable is on its own.

For example, in solving \(6a + 21 = 3a + 12\), subtraction of \(3a\) and \(21\) from either side results in \(3a = -9\), hence \(a = -3\) upon dividing by 3. It’s simple arithmetic steps that lead to finding the values that turn an equation to reality.

Algebraic Expressions

Algebraic expressions form the backbone of equations, with absolute values and linear equations both being types of expressions. In its simpler structure, it combines variables and constants with operations like addition and multiplication. When an absolute value is included, this alters its functionality:

• Represents distance, turning negative outcomes to positive, impacting how you solve equations.
• Often rewritten without the absolute, requiring careful polynomial manipulations.

Through conceptual understanding, a term like \(2a + 7\) can behave differently based on whether it’s evaluated within or outside absolute value signs. Part of algebraic expression solving within this exercise is recognizing when to rewrite and split, turning complex forms into simpler, manageable linear equations. This insight transforms intimidating math puzzles into a flow of solvable operations.