Problem 52 - Algebra 2

Question

Solve each equation. Check your solution. $$ \frac{1}{9}-\frac{2}{3} b=\frac{1}{18} $$

Step-by-Step Solution

Answer

The solution is \( b = \frac{1}{12} \).

Step 1: Understand the Equation

The given equation is \( \frac{1}{9} - \frac{2}{3}b = \frac{1}{18} \). We need to solve for \( b \).

Step 2: Eliminate Fractions

To eliminate the fractions, find the least common multiple (LCM) of the denominators 9, 3, and 18, which is 18. Multiply every term in the equation by 18 to remove the fractions:\[ 18 \left( \frac{1}{9} \right) - 18 \left( \frac{2}{3}b \right) = 18 \left( \frac{1}{18} \right) \] Simplifying each term, we have:\[ 2 - 12b = 1 \]

Step 3: Isolate the Variable Term

Get the \(-12b\) term by itself by subtracting 2 from both sides of the equation:\[ -12b = 1 - 2 \] Which simplifies to:\[ -12b = -1 \]

Step 4: Solve for the Variable

Divide both sides by -12 to solve for \( b \):\[ b = \frac{-1}{-12} \] Simplifying the right side gives:\[ b = \frac{1}{12} \]

Step 5: Check the Solution

Substitute \( b = \frac{1}{12} \) back into the original equation to verify:\[ \frac{1}{9} - \frac{2}{3} \left( \frac{1}{12} \right) \overset{?}{=} \frac{1}{18} \]The second term simplifies:\[ \frac{2}{3} \times \frac{1}{12} = \frac{1}{18} \]Substituting back, we have:\[ \frac{1}{9} - \frac{1}{18} = \frac{2}{18} - \frac{1}{18} = \frac{1}{18} \]Thus, the original equation holds true.

Key Concepts

Eliminate Fractions

Fractions can be intimidating when solving equations. It's like trying to juggle while riding a skateboard!
To make life easier, we can eliminate these fractions right from the start. This helps us work with simpler numbers instead.
Consider the equation you've encountered:

• \( \frac{1}{9} - \frac{2}{3} b = \frac{1}{18} \)

To eliminate fractions, you can multiply every term in the equation by a number that all denominators can divide into. This is called a common multiple. In this case, we will use the smallest one, called the least common multiple (LCM). Let's dive into understanding what LCM is next!

Least Common Multiple

The least common multiple (LCM) is the smallest number that is a multiple of each of the denominators in the equation. This may sound complex, but it’s just about finding a number that all your denominators share when multiplied.
In our equation, the denominators are 9, 3, and 18.

• Find the multiples of each denominator.
• 9: 9, 18, 27, 36...
• 3: 3, 6, 9, 12, 15, 18...
• 18: 18, 36, 54...

As you can see, 18 is the smallest number that appears in all lists of multiples.
It’s like finding a common dance floor for everyone to groove on! Multiplying each term by this LCM, you effectively clear out those fractions. So, multiply every part of \( \frac{1}{9} - \frac{2}{3}b = \frac{1}{18} \) by 18. Give it a try and see those fractions disappear!

Isolate Variable

Once fractions are out of the picture, you can easily focus on isolating the variable.
Imagine you're on a mission to uncover the secret treasure that is the value of \( b \). To do this, you need to get \( b \) all by itself on one side of the equation.
After multiplication, your equation becomes:

• \( 2 - 12b = 1 \)

To isolate \( b \), start by moving constant terms to the other side by subtracting 2 from both sides:

• \( -12b = -1 \)

Now, divide each side by \(-12\) to solve for \( b \):

• \( b = \frac{-1}{-12} \)

Simplifying, this gives \( b = \frac{1}{12} \)! You did it! You found the treasure.

Check Solution

After solving for the variable, it's crucial to check your work. Imagine you just finished baking a cake. Checking is like tasting a piece to make sure it's just right.
Substitute \( b = \frac{1}{12} \) back into the original equation to verify:

• \( \frac{1}{9} - \frac{2}{3} \left( \frac{1}{12} \right) \overset{?}{=} \frac{1}{18} \)

Evaluate the second term:

• \( \frac{2}{3} \times \frac{1}{12} = \frac{1}{18} \)

So substituting back, here's what happens:

• \( \frac{1}{9} - \frac{1}{18} = \frac{2}{18} - \frac{1}{18} = \frac{1}{18} \)

The left side equals the right side, confirming our solution is correct.
Rejoice, as the cake—or in this case, our equation—turned out perfect!