Problem 29 - College Algebra with Modeling and Visualization

Question

Solve the equation and check your answer. $$ \frac{2}{7} n+\frac{1}{5}=\frac{4}{7} $$

Step-by-Step Solution

Answer

The solution is \( n = \frac{13}{10} \).

Step 1: Isolate the variable term

To solve the equation, we first need to isolate the variable term \( \frac{2}{7}n \). We do this by subtracting \( \frac{1}{5} \) from both sides of the equation: \[\frac{2}{7}n + \frac{1}{5} - \frac{1}{5} = \frac{4}{7} - \frac{1}{5}\] This simplifies to: \[\frac{2}{7}n = \frac{4}{7} - \frac{1}{5}\]

Step 2: Subtract fractions on the right-hand side

Next, we need to subtract the fractions on the right-hand side of the equation. To do this, we find a common denominator. The least common multiple of 7 and 5 is 35:\[\frac{4}{7} = \frac{20}{35}\] \[\frac{1}{5} = \frac{7}{35}\] Now subtract:\[\frac{20}{35} - \frac{7}{35} = \frac{13}{35}\] Thus, the equation becomes:\[\frac{2}{7}n = \frac{13}{35}\]

Step 3: Solve for the variable

To solve for \( n \), we need to isolate \( n \) by multiplying both sides of the equation by the reciprocal of \( \frac{2}{7} \), which is \( \frac{7}{2} \):\[ n = \frac{13}{35} \times \frac{7}{2}\] Simplify:\[ n = \frac{13 \times 7}{35 \times 2} = \frac{91}{70}\] Simplify further:\[ n = \frac{13}{10} \]

Step 4: Verify the solution

To verify the solution, substitute \( n = \frac{13}{10} \) back into the original equation and see if both sides equal:\[ \frac{2}{7} \times \frac{13}{10} + \frac{1}{5} = \frac{4}{7} \] Calculate \( \frac{2}{7} \times \frac{13}{10} \):\[ \frac{26}{70} = \frac{13}{35} \]Add \( \frac{1}{5} \):\[ \frac{13}{35} + \frac{7}{35} = \frac{20}{35} = \frac{4}{7} \] Both sides are equal, confirming our solution.

Key Concepts

Fraction Operations

Fractions can often appear daunting but understanding their basic operations can simplify many problems. When working with fractions, it's important to remember that they are just numbers represented as ratios. Key operations involve addition, subtraction, multiplication, and division. For addition and subtraction, like in our problem, we need a common denominator. A common denominator is a shared multiple of the fractions' denominators. In our example, we converted \( \frac{4}{7} \) and \( \frac{1}{5} \) to have a common denominator of 35:

• \( \frac{4}{7} \rightarrow \frac{20}{35} \)
• \( \frac{1}{5} \rightarrow \frac{7}{35} \)

This conversion helps us to easily subtract the fractions:

• \( \frac{20}{35} - \frac{7}{35} = \frac{13}{35} \)

Understanding fraction operations allows us to transition smoothly to solving equations that involve these number forms.

Variable Isolation

Isolating the variable is one of the key steps in solving linear equations, as seen in this example. The goal is to "solve" the equation, which means finding the value of the variable that makes the equation true. For this, we want to have the variable term alone on one side of the equation. This is accomplished by performing operations that reverse what is being done to the variable.
In our problem, this means getting rid of the \( \frac{1}{5} \) term by subtracting it from both sides, leaving:

• \( \frac{2}{7}n = \frac{4}{7} - \frac{1}{5} \)

Once we have \( n \) as close as possible to being alone, we use the reciprocal of \( \frac{2}{7} \), which is \( \frac{7}{2} \), to finally isolate the variable:

• \( n = \frac{13}{35} \times \frac{7}{2} \)
• \( n = \frac{91}{70} = \frac{13}{10} \)

Once isolated, the variable can be solved for its value, which provides the solution to the equation.

Equation Verification

Verifying an equation ensures that your solution is accurate. This is a crucial step because it confirms whether the value you found for the variable truly satisfies the original equation. In our example, after solving for \( n \), substituting back into the original equation checks our work:

• Original: \( \frac{2}{7}n + \frac{1}{5} = \frac{4}{7} \)
• Substitute: \( \frac{2}{7} \times \frac{13}{10} + \frac{1}{5} \)
• Calculate: \( \frac{26}{70} + \frac{7}{35} \)

This confirms if both sides equal \( \frac{4}{7} \), proving the solution is correct. Here, they indeed matched, showing the equation balances perfectly. Verification prevents small errors from affecting larger mathematical problems. Always remember this validation step to ensure your solutions are precise and reliable.