Problem 102 - Intermediate Algebra

Question

Solve each equation. $$ \frac{2}{3}(2 x+2)+4=\frac{1}{6}(5 x+29) $$

Step-by-Step Solution

Answer

The solution to the equation is \( x = -1 \).

Step 1: Eliminate Fractions

To eliminate the fractions, we multiply every term by the least common multiple (LCM) of the denominators, which is 6. This gives us:\[ 6 \left( \frac{2}{3}(2x+2) + 4 \right) = 6 \left( \frac{1}{6}(5x+29) \right) \]This simplifies to:\[ 4(2x+2) + 24 = 5x + 29 \]

Step 2: Distribute and Simplify

Now, we'll distribute the 4 into the parentheses on the left side of the equation:\[ 8x + 8 + 24 = 5x + 29 \]Combine like terms on the left:\[ 8x + 32 = 5x + 29 \]

Step 3: Isolate the Variable

Subtract \( 5x \) from both sides to get all the \( x \)-terms on one side:\[ 8x - 5x + 32 = 29 \]This simplifies to:\[ 3x + 32 = 29 \]

Step 4: Solve for x

Subtract 32 from both sides to isolate the terms with \( x \):\[ 3x = 29 - 32 \]Simplify the right side:\[ 3x = -3 \]Divide each side by 3 to solve for \( x \):\[ x = -1 \]

Key Concepts

Fractions

Solving equations with fractions can seem tricky at first, but it becomes straightforward once you understand how to eliminate them. When you encounter fractions in an equation, the first step is to remove them by determining the least common multiple (LCM) of the denominations. This involves:

• Identifying the different denominators in the equation.
• Finding the smallest common multiple that each denominator divides into without a remainder.
• Multiplying every term in the equation by this LCM to eliminate the fractions.

By doing so, you transform the equation into one with whole numbers, making it easier to solve. For example, in the given exercise, the LCM of the numbers 3 and 6 is 6. Multiplying through by 6 eliminates the fractions, simplifying your work greatly.
Remember, clearing fractions makes it easier to focus on solving the equation itself, without the complication of fractional numbers.

Distributing Terms

After clearing fractions, the next important step is distributing terms. "Distributing" means multiplying each term inside a parenthesis by the number outside. This eliminates parentheses and simplifies the expression.

• Look for any group of terms within parentheses, with an outside factor.
• Multiply the outer factor across each term inside the parentheses.
• Simplify the resulting expression.

In our example, we distributed 4 over the terms inside the parenthesis like this:\[ 4(2x+2) = 8x + 8 \]By using distribution, you effectively dismantle expressions, preparing them for simpler operations. It's important to correctly distribute without missing any terms or factors, as this ensures the integrity of your equation.

Combining Like Terms

Once distribution is complete, the next step is combining like terms to further simplify the equation. "Like terms" are terms that contain the same variables raised to the same power. Here's how you simplify by combining them:

• Identify terms that have identical variable parts, such as terms with \( x \).
• Sum or subtract the coefficients of these terms directly.
• Combine constants in a similar manner.

This step makes the equation easier to solve as it minimizes the number of terms you have to deal with. In our solution, we combined terms on the left-hand side:\[ 8x + 8 + 24 \] which simplifies to \[ 8x + 32 \].
Effective combining streamlines the equation, setting the stage for isolating variables in the subsequent step.

Isolating Variables

The final step in solving the equation is isolating the variable to find its value. This involves rearranging the equation such that the variable is on one side of the equation, and constant terms are on the other.

• Start by getting all terms with the variable on one side. In our case, subtract \( 5x \) from both sides:
• Solve for the variable by performing opposite operations to "undo" any addition, subtraction, multiplication, or division it's subjected to.
• For example, subtract any constant added to the variable's side, and divide by the coefficient of the variable.

Applying these steps, we isolated \( x \):\[ 3x + 32 = 29 \]Subtract 32 from both sides:\[ 3x = -3 \]Then divide by 3:\[ x = -1 \].
Isolating variables is crucial as it allows you to explicitly solve for the unknown. The property of operations ensures balance and maintains equality throughout your manipulations.