Problem 26 - Intermediate Algebra

Question

Solve each proportion. $$ \frac{9 t+6}{t}=\frac{7}{3} $$

Step-by-Step Solution

Answer

The solution is \( t = \frac{-9}{10} \).

Step 1: Cross-Multiply

To solve a proportion, we start by cross-multiplying the terms. This involves multiplying the numerator of one fraction by the denominator of the other fraction.Set up the equation: \[ 3(9t + 6) = 7t \]

Step 2: Distribute

Next, distribute the 3 across the terms inside the parentheses.Distribute 3:\[ 3 imes 9t + 3 imes 6 = 7t \]Simplify:\[ 27t + 18 = 7t \]

Step 3: Move Terms to One Side

We need to get all terms with \( t \) on one side of the equation and constants on the other. So, subtract \( 7t \) from both sides to isolate \( t \).Subtract \( 7t \) from both sides:\[ 27t - 7t + 18 = 7t - 7t \]Simplify:\[ 20t + 18 = 0 \]

Step 4: Isolate the Variable

Subtract 18 from both sides to further isolate the term with \( t \).Subtract 18:\[ 20t + 18 - 18 = 0 - 18 \]Simplify:\[ 20t = -18 \]

Step 5: Solve for t

Finally, solve for \( t \) by dividing both sides by 20.\[ t = rac{-18}{20} \]Simplify the fraction by dividing both the numerator and the denominator by 2:\[ t = rac{-9}{10} \]

Key Concepts

Cross-Multiplication

Cross-multiplication is an essential first step in solving proportions. When you have an equation where two fractions are set equal to each other, you can eliminate the fractions by cross-multiplying. This means you take the numerator of one fraction and multiply it by the denominator of the other fraction, and do the same with the remaining numerator and denominator.

In our example, the proportion is \[ \frac{9t+6}{t} = \frac{7}{3} \]. Here, you multiply the 3 by the numerator on the left, which is \(9t+6\), and multiply 7 by the denominator \(t\).

• Set up the cross-multiplied equation: \[ 3(9t+6) = 7t \]

Cross-multiplication helps us condense the equation to a form where easier algebraic manipulation like distributing and simplifying can occur.

Distributing Terms

Once you have executed cross-multiplication, the next task is to distribute terms. Distribution involves multiplying a single term by each term inside parentheses.

In our equation \[ 3(9t + 6) = 7t \], the 3 must be distributed across the terms within the parentheses:

• Multiply 3 by \(9t\): \(3 \times 9t = 27t\)
• Multiply 3 by 6: \(3 \times 6 = 18\)

This distribution results in the equation: \[ 27t + 18 = 7t \]. Distributing terms effectively breaks down components within an equation, paving the way for easier manipulation and isolation of variables.

Isolating Variables

After distributing terms, the goal is to isolate the variable, usually on one side of the equation. This means getting all terms involving the variable on one side and constants on the other.

Our current equation is: \[ 27t + 18 = 7t \]. You can subtract \(7t\) from both sides to bring all\(t\) terms together:

• Subtract \(7t\) from the left side: \(27t - 7t\) results in \(20t\)
• The equation becomes \(20t + 18 = 0\)

Next, we need to isolate \(t\) further by getting it alone. Subtract18 from both sides:

• \(20t + 18 - 18 = 0 - 18\)
• This simplifies to \(20t = -18\)

Isolating variables is crucial, as it moves us closer to finding the value of \(t\) by ensuring all terms are properly organized.

Fractions Simplification

In the final steps of solving equations, you often need to simplify fractions. Simplifying fractions involves reducing them to their simplest form by dividing the numerator and denominator by their greatest common divisor.

The current equation \[ 20t = -18 \]can be rewritten to solve for \(t\) by dividing both sides by 20:

• \(t = \frac{-18}{20}\)

This fraction can be simplified further. Both 18 and 20 can be divided by 2, their greatest common divisor:

• Divide the numerator by 2: \(-18 \div 2 = -9\)
• Divide the denominator by 2: \(20 \div 2 = 10\)

Resulting in the simplified solution:\[ t = \frac{-9}{10} \]. By simplifying fractions, you ensure your solution is as neat and exact as possible. Reducing fractions helps in achieving precision and clarity in mathematical solutions.