Problem 108 - Introductory Algebra for College Students

Question

Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. It takes me more steps to find \(\frac{2}{x+5}+\frac{7}{x-3}\) than it does to find \(\frac{2 x^{3}}{x+5}+\frac{7 x^{3}}{x+5}\)

Step-by-Step Solution

Answer

The statement does not make sense, the second expression involves more steps to solve.

Step 1: Analyze the first expression

Look at the first expression \(\frac{2}{x+5}+\frac{7}{x-3}\). This expression can be solved much simply as it only involves adding fractions, which involve fewer steps.

Step 2: Analyze the second expression

Now look at the second expression \(\frac{2 x^{3}}{x+5}+\frac{7 x^{3}}{x+5}\). It involves multiplication of fractions and addition, introducing more steps.

Step 3: Compare both expressions

Comparing both expressions, it becomes clear that the second expression requires more steps to solve, as it involves multiplication or possibly additional simplification steps if \(x\) has a specified value.

Step 4: Conclude

So, the provided statement 'It takes me more steps to find \(\frac{2}{x+5}+\frac{7}{x-3}\) than it does to find \(\frac{2 x^{3}}{x+5}+\frac{7 x^{3}}{x+5}\)' does not make sense, because the latter expression requires more computational steps.

Key Concepts

Addition of Fractions

Adding fractions may seem tricky at first, especially when they have different denominators. Let's break it down. When you add fractions, you need to have a common denominator. This is the number on the bottom of the fraction.
When fractions have different denominators, like \( \frac{2}{x+5} \) and \( \frac{7}{x-3} \), you can't just add the numbers on the top.
You have to find a common denominator by multiplying the denominators together, or finding the least common multiple.

• First, determine the common denominator, which is the product \((x+5)(x-3)\).
• Next, adjust the fractions to have this common denominator.
• For \(\frac{2}{x+5}\), multiply both numerator and denominator by \((x-3)\) to get \(\frac{2(x-3)}{(x+5)(x-3)}\).
• For \(\frac{7}{x-3}\), multiply by \((x+5)\) to get \(\frac{7(x+5)}{(x+5)(x-3)}\).

Now, you are ready to add the numerators: \(2(x-3) + 7(x+5)\). Just remember, the denominator stays the same.
Simplifying the numerators gives the result, all while maintaining the combined denominator.

Multiplication of Fractions

Multiplying fractions is a straightforward process compared to addition. Unlike addition, you do not need a common denominator when multiplying.
All you have to do is multiply the numerators together and then the denominators together.
In the expression \( \frac{2x^3}{x+5} + \frac{7x^3}{x+5} \), notice how both fractions already have a common denominator, which simplifies things. However, due to the addition involved, one might get confused.

• Since multiplication doesn't require a common denominator, simplify the terms as needed before multiplying.
• Sometimes, important simplifications can occur before multiplying, such as factoring out common terms.

In the above expression, notice both numerators, \(2x^3\) and \(7x^3\), share a common factor of \(x^3\). Multiplying like terms can help in simplifying the end results. After understanding the multiplication, refer back to addition, since this expression also involves adding two fractions. The key here is simplifying before executing multiplication and addition to make the process more efficient.

Simplification in Algebra

Simplification is a key skill in algebra that makes complex expressions easier to understand and solve.
When you're faced with expressions like those in our exercise, it helps to always look for ways to reduce the number of terms or simplify the components.
In the exercise provided, simplifying plays a huge part in computing the second expression.

• Identify common factors in the expression. For example, \(2x^3\) and \(7x^3\) share \(x^3\) as a common factor.
• Combine like terms, just as you would with simpler numerical expressions.
• Factoring is an essential tool, as it often reveals simpler paths to the solution.

Simplifying fractions, whether during addition, subtraction, multiplication, or division, helps speed up calculations.
Reducing everything to the simplest form is the standalone way of ensuring accuracy in solving complex algebraic problems.
Always aim for making expressions as manageable as possible before moving forward in solving them.