Problem 10 - Elementary Algebra

Question

For Problems \(1-12\), solve each of the equations. These equations are the types you will be using in Problems 13-40. $$ \frac{10}{3} r+\frac{10}{3}(r-3)=90 $$

Step-by-Step Solution

Answer

The solution is \( r = 15 \).

Step 1: Expand the expression

The given equation is \( \frac{10}{3} r + \frac{10}{3}(r-3) = 90 \). To start, distribute the \( \frac{10}{3} \) across the parentheses. This gives us \( \frac{10}{3} r + \frac{10}{3} r - \frac{10}{3} \times 3 = 90 \). Simplifying the term \( \frac{10}{3} \times 3 \) results in \( 10 \). Thus, the equation becomes \( \frac{10}{3} r + \frac{10}{3} r - 10 = 90 \).

Step 2: Combine like terms

Now that the expression is expanded, combine like terms. We have \( \frac{10}{3} r + \frac{10}{3} r \), which is \( \frac{20}{3} r \). The equation then becomes \( \frac{20}{3} r - 10 = 90 \).

Step 3: Isolate the variable term

Add 10 to both sides of the equation to isolate the term with the variable \( r \): \( \frac{20}{3} r = 90 + 10 \). Simplifying the right side, we get \( \frac{20}{3} r = 100 \).

Step 4: Solve for \( r \)

To solve for \( r \), multiply both sides of the equation by the reciprocal of \( \frac{20}{3} \), which is \( \frac{3}{20} \). Performing this operation, we get \( r = 100 \times \frac{3}{20} \). Simplifying, \( 100 \times \frac{3}{20} = 15 \). Thus, \( r = 15 \).

Key Concepts

Distributive Property

The distributive property is a fundamental algebraic principle used to simplify equations. It states that a term multiplied by an expression in parentheses can be expanded by multiplying each term inside the parentheses by that outside term.

Let's consider the equation:

• \(\frac{10}{3}(r-3)\)

Here, the term \(\frac{10}{3}\) is distributed across \(r-3\). This means you multiply \(\frac{10}{3}\) by \(r\) and also by \(-3\). So you get:

• \(\frac{10}{3} \cdot r\) and \(-\frac{10}{3} \cdot 3\)

This action helps by breaking down more complex expressions into simpler parts. The result from this step sets the stage for solving the equation by providing a clear view of all the terms involved.

Combining Like Terms

After distributing terms in an equation, it is crucial to combine like terms in order to simplify further. Like terms are terms that contain the same variable raised to the same power.

In our particular equation, after using the distributive property, we end up with:

• \(\frac{10}{3}r + \frac{10}{3}r - 10\)

Since both \(\frac{10}{3}r\) terms contain 'r', they are like terms. By combining them, you perform addition and get:

• \(\frac{20}{3}r\)

Combining like terms simplifies the equation, making it more straightforward and easier to solve. This is a critical step as it brings us closer to isolating the variable and solving for its value.

Isolating the Variable

One of the primary goals in solving linear equations is isolating the variable, which means getting the variable on one side of the equation. This allows you to solve for its value directly.

In the equation \(\frac{20}{3}r - 10 = 90\), start by eliminating constant terms on the side of the equation with the variable by adding or subtracting them from both sides.

In this case:

• Add 10 to both sides to cancel out the \(-10\):
  - \(\frac{20}{3}r - 10 + 10 = 90 + 10\)
  - Simplifies to \(\frac{20}{3}r = 100\)

This step effectively removes obstacles around the variable, setting up for the final solve step. Ensuring the variable is isolated paves the way for precise calculation of its value.

Reciprocal of a Fraction

When dealing with equations involving fractions, multiplying by a reciprocal is a key technique to solve for the variable. A reciprocal simply flips a fraction, transforming the numerator into the denominator and vice versa.

If we look at the equation \(\frac{20}{3}r = 100\), to solve for \(r\), we can multiply both sides of the equation by the reciprocal of \(\frac{20}{3}\), which is \(\frac{3}{20}\).

• \(r = 100 \times \frac{3}{20}\)
• Simplifying the multiplication: \(r = 15\)

Multiplying by the reciprocal effectively "cancels out" the coefficient of the variable, leaving you with the variable alone. This step is crucial to find the exact value of the variable, hence completing the solution of the equation.