Problem 19 - Algebra 2 and Trigonometry

Question

In \(3-20,\) solve each equation and check. $$ \frac{4}{y+2}=1-\frac{8}{y(y+2)} $$

Step-by-Step Solution

Answer

The solution is \( y = 4 \).

Step 1: Identify and Combine Fractions

First, observe that the equation is \( \frac{4}{y+2} = 1 - \frac{8}{y(y+2)} \). To make it easier to solve, let's combine all the terms to have a common denominator. The common denominator here is \( y(y+2) \). Rewrite each term with this denominator.

Step 2: Simplify the Left Side

Multiply the left side by \( \frac{y}{y} \) to obtain \( \frac{4y}{y(y+2)} \). Now the left side \( \frac{4}{y+2} \) is expressed with the common denominator.

Step 3: Simplify the Equation

The equation now is \( \frac{4y}{y(y+2)} = \frac{y(y+2)}{y(y+2)} - \frac{8}{y(y+2)} \). The right side simplifies to \( \frac{y^2 + 2y - 8}{y(y+2)} \) by recognizing that \( 1 = \frac{y(y+2)}{y(y+2)} \).

Step 4: Equate the Numerators

Since the denominators are the same, set the numerators equal: \( 4y = y^2 + 2y - 8 \). This is now a quadratic equation.

Step 5: Rearrange into Standard Quadratic Form

Rearrange the equation \( y^2 + 2y - 8 = 4y \) to \( y^2 - 2y - 8 = 0 \) by subtracting \( 4y \) from both sides.

Step 6: Factor the Quadratic Equation

Factor the quadratic equation \( y^2 - 2y - 8 = 0 \). The factors are \((y - 4)(y + 2) = 0\).

Step 7: Solve for y

Solve \( (y - 4)(y + 2) = 0 \). This gives \( y = 4 \) or \( y = -2 \).

Step 8: Check the Solutions

Substitute \( y = 4 \) back into the original equation: \( \frac{4}{4+2} = 1 - \frac{8}{4(4+2)} \). This simplifies to \( \frac{4}{6} = 1 - \frac{1}{3} \), which is a true statement.Substitute \( y = -2 \) back into the original equation: The left side yields \( \frac{4}{0} \), which is undefined, so \( y = -2 \) is not a valid solution.

Key Concepts

Quadratic Equations

Quadratic equations are mathematical expressions that take the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations can have up to two solutions, called the roots of the equation.
To solve quadratic equations, you have several techniques at your disposal:

• Factoring: This is the method of expressing the quadratic equation as a product of two binomials, such as \((x - p)(x - q) = 0\). If the product equals zero, then either \(x - p = 0\) or \(x - q = 0\), giving us the roots \(x = p\) and \(x = q\).
• Completing the square: This technique involves rewriting the equation in the form \((x - h)^2 = k\) to find the solutions.
• Quadratic Formula: By using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), we can find the roots of any quadratic equation, even if they aren't easily factorable.

In our given equation, after combining terms, the resulting quadratic was \(y^2 - 2y - 8 = 0\). We then used factoring to find its roots: \(y = 4\) and \(y = -2\). However, after checking, we realized that only \(y = 4\) is valid.

Factoring Polynomials

Factoring polynomials is a crucial skill that makes solving equations much simpler. It involves breaking down a polynomial into the product of its simpler polynomials. When we factor, we're essentially "unfolding" the equation to understand its roots.
In the quadratic equation \(y^2 - 2y - 8\), to factor it, we look for two numbers that multiply to give \(-8\) (the constant term) and add up to \(-2\) (the coefficient of \(y\)). The numbers \(-4\) and \(2\) fit the bill, allowing us to express the equation as

• \((y - 4)(y + 2) = 0\)

By setting each factor to zero, we solve for \(y\):

• \(y - 4 = 0 \implies y = 4\)
• \(y + 2 = 0 \implies y = -2\)

These techniques of recognizing factor pairs are invaluable when solving quadratic equations, making the process faster and more intuitive.

Common Denominators

When dealing with fractions, especially in equations like the one we solved \( \frac{4}{y+2} = 1 - \frac{8}{y(y+2)} \), using common denominators is key. This practice allows us to combine fractions easily, simplifying the equation to a more manageable form.
The common denominator is a shared multiple of the denominators in each fraction. In the exercise, this was \(y(y+2)\), which is the least common multiple of \(y+2\) and \(y(y+2)\).
This approach involves first rewriting each term in the equation so that all terms share the same denominator. Here’s how it works:

• Multiply the numerator and denominator of \(\frac{4}{y+2}\) by \(y\) to get \(\frac{4y}{y(y+2)}\).
• The expression \(1\) is rewritten with the common denominator, becoming \(\frac{y(y+2)}{y(y+2)}\).

Now, all terms on both sides of the equation have this common denominator, allowing us to eliminate the denominators and focus on solving the equation.

Undefined Expressions

In math, undefined expressions occur when we run into situations that lack a clear numerical value. A common scenario is division by zero. In our exercise, when we attempted the solution \(y = -2\), we saw this in action. The expression \(\frac{4}{y+2}\) becomes \(\frac{4}{0}\) when \(y = -2\), which is undefined because division by zero is not possible.
Undefined expressions pose significant problems as they halt the process of solving an equation. Mathematically, an operation must always produce a number or a well-defined value; anything else signals a mistake or unviable solution.
When solving equations, it’s essential to remember:

• Check for values where the denominator equals zero.
• Eliminate or rule out such values as potential solutions.

In conclusion, while solving our problem, we found that although \(y = -2\) algebraically appears to be a solution, it is not valid because it results in an undefined mathematical operation.