Problem 44 - Elementary Algebra

Question

For the following problems, solve the rational equations. $$ \frac{6}{y^{2}}-\frac{5}{y}=1 $$

Step-by-Step Solution

Answer

The solutions for the rational equation are \(y = -6\) and \(y = 1\).

Step 1: Find the common denominator

In this case, the denominators are \(y^2\) and \(y\). The common denominator (also known as Least Common Denominator or LCD) is the smallest power of y that both denominators can divide into. In this case, it is \(y^2\).

Step 2: Eliminate fractions by multiplying each term by the common denominator

We will multiply the whole equation by the common denominator, \(y^2\). This will allow us to get rid of the fractions in the equation. $$ y^2\left(\frac{6}{y^{2}}-\frac{5}{y}\right) = y^2(1) $$

Step 3: Simplify the equation

After multiplying, we will simplify the equation by canceling out common factors. In this case, we have: $$ 6y^2\left(\frac{1}{y^{2}}\right)-5y\left(\frac{y^2}{y}\right)= y^2 $$ Which simplifies to: $$ 6-5y = y^2 $$ Now we have a quadratic equation.

Step 4: Move all terms to one side of the equation

In order to solve the quadratic equation, we need to have all the terms on one side of the equation and set the equation equal to zero. In this case, we will move the terms to the left side of the equation: $$ y^2+5y-6 = 0 $$

Step 5: Factor the quadratic equation

Now, we will factor the quadratic equation: $$ (y+6)(y-1) = 0 $$

Step 6: Solve for y

Finally, we can solve for y by setting each factor equal to zero: $$ y+6 = 0 \Rightarrow y = -6 $$ and $$ y-1 = 0 \Rightarrow y= 1 $$ So the two solutions are \(y = -6\) and \(y = 1\).

Key Concepts

Understanding the Common Denominator

When solving rational equations, finding a common denominator is crucial. It serves as a foundation for combining fractions, allowing us to simplify more complicated expressions. The common denominator is the smallest expression that each denominator can divide into without leaving a remainder.

In the example problem, our denominators are \(y^2\) and \(y\). The smallest common denominator that accommodates both \(y\) and \(y^2\) is \(y^2\).

• This choice ensures that both original denominators can evenly divide into the chosen denominator.
• Using a common denominator simplifies the solution process by giving us a single expression to work with, making manipulation of the equation easier.

This step is essential before you can remove fractions and proceed with other algebraic operations like multiplication.

Transforming into a Quadratic Equation

Once we've identified and used a common denominator, the next step is transforming the equation so that it's simpler to manage. This involves eliminating the fractions by multiplying the entire equation by the common denominator, which in our case is \(y^2\).

Through this process, each term can be rewritten without fractions, resulting in an equation like \(6 - 5y = y^2\).

• With the fractions gone, we can more easily reconfigure the equation into standard quadratic form: \(ax^2 + bx + c = 0\).
• This transformation is a pivotal part of solving rational equations, setting us up for the subsequent steps of manipulation and solution.

In our example, rearranging leads to \(y^2 + 5y - 6 = 0\), a classic quadratic equation.

Factoring the Quadratic Equation

Factoring is a powerful algebraic tool that can help solve quadratic equations. It involves expressing a polynomial as a product of simpler, more manageable terms. In the context of our rational equation, once reduced to \(y^2 + 5y - 6 = 0\), we identify two numbers whose product is \(-6\) (the constant term) and whose sum is \(5\) (the coefficient of \(y\)).

This leads us to the factors \((y + 6)\) and \((y - 1)\).

• Each factor represents a potential solution to the equation, where the product of the two is zero.
• Understanding how to factor effectively simplifies solving quadratic equations, making it easier to find the roots.

Factoring is particularly useful because it breaks down complex expressions into their component parts, which can then be analyzed individually.

Solving Rational Equations using Roots

The final step in solving rational equations involves working with the factored form to find possible solutions for the variable. After factoring \(y^2 + 5y - 6 = 0\) into \((y + 6)(y - 1) = 0\), we use each factor to solve for \(y\).

By setting each factor equal to zero, we find two potential solutions: \(y + 6 = 0\) giving \(y = -6\) and \(y - 1 = 0\) giving \(y = 1\).

• Solving these simple equations reveals the roots or solutions of the original quadratic equation.
• These solutions are checked back into the context of the original problem to ensure they make sense within any constraints given by the rational expression.

It's important to emphasize solving these simple equations carefully, as each root provides insight into where the original rational equation holds true.