Problem 52 - Introductory Algebra for College Students

Question

Solve or simplify, whichever is appropriate. $$3 y^{-2}+1=4 y^{-1}$$

Step-by-Step Solution

Answer

The solutions for the given equation are \(y = 3\) and \(y = 1\).

Step 1: Transform the equation

Transform the equation by multiplying every term by \(y^2\), to get rid of the exponents. This results in \(3 + y^2 = 4y\).

Step 2: Rearrange the equation

Rearrange the equation to set it equal to zero, resulting in \(y^2 - 4y + 3 = 0\).

Step 3: Factor the quadratic

Factor the quadratic equation to solve for 'y'. This results in \((y - 3)(y - 1) = 0\).

Step 4: Solve for y

Setting each factor equal to zero gives the solutions \(y = 3\) and \(y = 1\).

Key Concepts

Factoring Quadratics

To solve quadratic equations like \(y^2 - 4y + 3 = 0\), factoring is a helpful method. Quadratic equations often take the form \(ax^2 + bx + c = 0\). The goal is to express this equation as a product of two binomials. In our case, we're looking to factor \(y^2 - 4y + 3\).

We need to find two numbers that multiply to the constant term (3) and add up to the linear coefficient (-4). These numbers are -3 and -1. Therefore, we can write:

• \( (y - 3)(y - 1) = 0 \)

Factoring sets the stage for solving the equation because if the product of two numbers is zero, at least one of them must be zero. Hence, we set each binomial equal to zero to find:

• \( y - 3 = 0 \Rightarrow y = 3 \)
• \( y - 1 = 0 \Rightarrow y = 1 \)

These are our solutions!

Rational Exponents

Rational exponents are expressions where variables or numbers are raised to a fraction as an exponent. They offer an alternative way of expressing roots. In the equation \(3 y^{-2}+1=4 y^{-1}\), the exponents \(-2\) and \(-1\) are negative rational exponents.

To handle rational exponents, it's important to remember these key points:

• A negative exponent indicates a reciprocal. For example, \(y^{-1} = \frac{1}{y}\).
• The expression \(y^{-2}\) can be rewritten as \(\frac{1}{y^2}\).

Rewriting the exponents in this way helps in transforming equations into more familiar forms without exponents, which we will cover in the next section.

Equation Transformation

Transforming equations is a key concept in making challenging problems more manageable. The initial step in our problem involved converting \(3 y^{-2} + 1 = 4 y^{-1}\) into a simpler form.

First, we cleared the negative exponents by multiplying every term by \(y^2\). This universal multiplication results in:

• \(y^2 \times 3 y^{-2} = 3\)
• \(y^2 \times 1 = y^2\)
• \(y^2 \times 4 y^{-1} = 4y\)

Thus, the equation simplifies to \(3 + y^2 = 4y\).

Next, we rearrange terms to bring all terms to one side, forming \(y^2 - 4y + 3 = 0\). With this new quadratic form, solving becomes easier through methods such as factoring, as explained previously. This way, equation transformation aids in systematically solving problems by changing the equation's form.