Problem 23 - Elementary Algebra

Question

For exercises 13-24, rewrite each expression as an equivalent expression with the given denominator. $$ \frac{3 c}{c^{2}-13 c+40} ;(c-8)(c-2)(c-5) $$

Step-by-Step Solution

Answer

\(\frac{3c^2 - 6c}{(c-8)(c-2)(c-5)}\)

Step 1 - Factor the Denominator

First, factor the quadratic expression in the denominator: \ \(c^2 - 13c + 40\). \ Look for two numbers that multiply to 40 and add to -13. These numbers are -5 and -8. So, the factorization is: \ \(c^2 - 13c + 40 = (c-8)(c-5)\).

Step 2 - Rewrite the Expression

Rewrite the original expression with the factored denominator: \ \(\frac{3c}{(c-8)(c-5)}\). \ We need to match the given denominator of \ \((c-8)(c-2)(c-5)\).

Step 3 - Find Equivalent Expression

Notice that the original denominator \ \((c-8)(c-5)\) is missing the factor \ \((c-2)\). To make the denominators match, multiply the numerator and the denominator by \ \((c-2)\): \ \(\frac{3c}{(c-8)(c-5)} * \frac{(c-2)}{(c-2)} = \frac{3c(c-2)}{(c-8)(c-2)(c-5)}\).

Step 4 - Simplify the Numerator

Distribute the term in the numerator: \ \(3c(c-2) = 3c^2 - 6c\). \ So the equivalent expression is: \ \(\frac{3c^2 - 6c}{(c-8)(c-2)(c-5)}\).

Key Concepts

Equivalent Expressions

In algebra, equivalent expressions are different ways of writing the same mathematical expression. They may look different but they mean the same thing.
Equivalent expressions are used to simplify and solve equations.

• Consider the expression \(\frac{3c}{c^2-13c+40}\).
• If we factor the denominator as shown in the solution, we get \(c^2-13c+40 = (c-8)(c-5)\).
• To match the given denominator \((c-8)(c-2)(c-5)\), we must multiply the numerator and the denominator by \(c-2\).
• This results in a new expression: \(\frac{3c \, (c-2)}{(c-8)(c-5)(c-2)}\).

These different forms are 'equivalent expressions', meaning they simplify to the same value for any valid value of \(c\).This equivalency allows for flexibility in problem-solving and simplifying fractions.

Common Denominators

A common denominator is a shared multiple of the denominators of two or more fractions. Having a common denominator is essential when adding, subtracting, or comparing fractions.

• For the exercise, the initial fraction is \(\frac{3c}{(c-8)(c-5)}\).
• We need to rewrite it with the new denominator \((c-8)(c-2)(c-5)\).
• To do this, multiply both the numerator and the denominator by the missing factor \(c-2\).

So, we get \( \frac{3c \, (c-2)}{(c-8)(c-5)(c-2)}\).
With this equivalent fraction, both the denominators and structures match, making further operations like addition or subtraction straightforward.

Algebraic Fractions

Algebraic fractions contain algebraic expressions in their numerator or denominator.
They follow the same rules as numeric fractions:

• Factor common terms.
• Find equivalent fractions.
• Simplify if possible.

In this problem, we start with \(\frac{3c}{c^2-13c+40}\).By factoring the denominator, we obtain components: \(c-8\) and \(c-5\).Next, to match the given denominator \(c-8)(c-2)(c-5)\), we multiply both the numerator and denominator by \(c-2\).Then, the fraction becomes \( \frac{3c \, (c-2)}{(c-8)(c-2)(c-5)}\).Finally, simplify the numerator by distributing: \(3c(c-2) = 3c^2 - 6c\).This results in the final expression: \( \frac{3c^2 - 6c}{(c-8)(c-2)(c-5)}\).Understanding algebraic fractions helps with simplifying complex equations and solving higher-level math problems.