Problem 57 - Elementary Algebra

Question

Write the following expressions using only positive exponents. Assume all variables are nonzero. $$ \left(5 y^{2}+8 y-6\right)^{-2}(6 y-1)^{-7} $$

Step-by-Step Solution

Answer

Question: Rewrite the expression \(\left(5 y^{2}+8 y-6\right)^{-2}(6 y-1)^{-7}\) with positive exponents. Answer: \(\frac{1}{(5y^2 + 8y - 6)^2 \cdot (6y - 1)^7}\)

Step 1: Rewrite the expression using the property \((a^{-n}) = \frac{1}{a^n}\)

We start by rewriting the given expression using the property \((a^{-n}) = \frac{1}{a^n}\): $$ \left(5 y^{2}+8 y-6\right)^{-2}(6 y-1)^{-7} = \frac{1}{(5y^2 + 8y - 6)^2} \cdot \frac{1}{(6y-1)^7} $$

Step 2: Combine the fractions

Now, we will combine the fractions by multiplying their numerators and denominators: $$ \frac{1}{(5y^2 + 8y - 6)^2} \cdot \frac{1}{(6y-1)^7} = \frac{1 \cdot 1}{(5y^2 + 8y - 6)^2 \cdot (6y - 1)^7} $$

Step 3: Simplify the expression

As the numerators are both 1, we can simplify the expression as follows: $$ \frac{1}{(5y^2 + 8y - 6)^2 \cdot (6y - 1)^7} $$ This is the final simplified expression with only positive exponents.

Key Concepts

Algebraic expressions

Algebraic expressions are combinations of numbers, variables, and mathematical operations like addition and multiplication. They can be as simple as a single number or variable, or as complex as a polynomial with multiple terms. In the given exercise, the expressions are \(5y^{2} + 8y - 6\) and \(6y - 1\). These represent polynomials, which are just sums of terms created by multiplying constants and variables raised to powers. Understanding algebraic expressions is fundamental because they form the basis for building and solving equations.
When dealing with expressions like \(5y^{2} + 8y - 6\), each component has its role. The \(y^2\) term indicates that the variable \(y\) is squared, and the coefficients (5 and 8) adjust the size of the terms when summed up. Simplifying such expressions usually involves rearranging them or applying arithmetic rules to make them easier to manipulate.

Exponent rules

Exponent rules are crucial for manipulating expressions involving powers. An important rule showcased in the exercise is \(a^{-n} = \frac{1}{a^n}\). This rule converts negative exponents into fractions with positive exponents. For example, in the original expression \(\left(5 y^{2}+8 y-6\right)^{-2}(6 y-1)^{-7}\), each part with a negative exponent is converted:

• \(\left(5 y^{2}+8 y-6\right)^{-2}\) becomes \(\frac{1}{(5 y^{2}+8y-6)^2}\).
• \((6y-1)^{-7}\) converts to \(\frac{1}{(6y-1)^7}\).

These transformations facilitate better manipulation and simplification of expressions.
Other exponent rules include the product of powers \(a^m \cdot a^n = a^{m+n}\) and the power of a power \( (a^m)^n = a^{m }\). These rules make it easier to handle polynomials during various algebraic operations, helping you keep expressions manageable and simplified.

Fraction simplification

Fraction simplification is the process of reducing expressions to their simplest form, which often involves factoring and canceling common terms. In the given solution, the original expression is rewritten using positive exponents:
The step-by-step reduction results in \((\frac{1}{(5 y^{2} + 8y - 6)^{2}}) \cdot (\frac{1}{(6y-1)^7})\). Combining these by multiplying the numerators and denominators gives:

• The new fraction \(\frac{1}{(5 y^{2} + 8y - 6)^{2} (6y - 1)^7}\).

Fractions are simplified by ensuring numerators and denominators contain no common factors. Since the example fraction's numerator is 1, only the denominator could potentially have factors simplifying it. No further steps are needed here as it’s already expressed in simplest form with positive exponents.
Mastering fraction simplification is essential when dealing with algebraic expressions, allowing students to present complex relations compactly and accurately.