Problem 20 - Precalculus Student Solutions Manual 5th

Question

Solve each equation. $$\frac{1}{15}(2 x+5)=\frac{x+2}{9}$$

Step-by-Step Solution

Answer

x = -5

Step 1: Clear the fractions by finding a common multiple

To eliminate the fractions, multiply both sides of the equation by the least common multiple (LCM) of the denominators 15 and 9. The LCM of 15 and 9 is 45. Thus, multiply both sides of the equation by 45: \[ 45 \times \frac{1}{15}(2x + 5) = 45 \times \frac{x + 2}{9} \]

Step 2: Simplify both sides

Distribute 45 on both sides of the equation: \[ 45 \times \frac{1}{15} (2x + 5) = 45 \times \frac{x + 2}{9} \]This simplifies to: \[ 3 \times (2x + 5) = 5 \times (x + 2) \]

Step 3: Distribute and simplify

Expand the terms on both sides by distributing the multiplication: \[ 3 \times 2x + 3 \times 5 = 5 \times x + 5 \times 2 \]Simplifying, we get: \[ 6x + 15 = 5x + 10 \]

Step 4: Solve for x

Isolate the variable, x, by first subtracting 5x from both sides: \[ 6x - 5x + 15 = 5x - 5x + 10 \]This simplifies to: \[ x + 15 = 10 \]Then, subtract 15 from both sides: \[ x + 15 - 15 = 10 - 15 \]Thus: \[ x = -5 \]

Key Concepts

least common multiple

To start solving the linear equation \(\frac{1}{15}(2x + 5) = \frac{x + 2}{9}\), we first need to get rid of the fractions. The key concept here is finding the least common multiple (LCM) of the denominators. \[ \text{Denominators: } 15, 9. \text{LCM: } 45 \] The LCM is the smallest number that both denominators can divide into without leaving a remainder. In this case, 45 is that number because both 15 and 9 can divide it evenly. Once we have the LCM, multiply both sides of the equation by 45: \[ 45 \times \frac{1}{15}(2x + 5) = 45 \times \frac{x + 2}{9} \] This step eliminates the fractions, making it easier to work with the equation.

distributive property

After multiplying both sides by the LCM, the next step involves using the distributive property. The distributive property states that \[ a(b + c) = ab + ac \] This property allows us to distribute the multiplication over the addition inside the parentheses. Let's apply it to our equation: \[ 45 \times \frac{1}{15}(2x + 5) = 3(2x + 5) = 6x + 15 \] Similarly, distribute on the other side: \[ 45 \times \frac{x + 2}{9} = 5(x + 2) = 5x + 10 \] After applying the distributive property, our equation becomes: \[ 6x + 15 = 5x + 10 \]

isolate variable

The final step is to isolate the variable, x, in the equation \[ 6x + 15 = 5x + 10 \] To isolate x, we start by getting all x-terms on one side of the equation and constant terms on the other. First, subtract 5x from both sides: \[ 6x - 5x + 15 = 5x - 5x + 10 \] Simplifying gives: \[ x + 15 = 10 \] Next, subtract 15 from both sides to solve for x: \[ x + 15 - 15 = 10 - 15 \] Thus: \[ x = -5 \] So, x is isolated and the solution to the equation is \text{x = -5.\} By understanding how to isolate the variable, you can simplify and solve linear equations seamlessly.