Problem 20 - Beginning and Intermediate Algebra

Question

Solve. $$\frac{3 y+5}{y^{2}+5 y}+\frac{y+4}{y+5}=\frac{y+1}{y}$$

Step-by-Step Solution

Answer

The short answer to the given equation is \(y = 0\) and \(y = \frac{1}{2}\).

Step 1: Finding the Common Denominator

The denominators of the fractions on the LHS are \(y^{2}+5 y\) and \(y+5\). The denominator on the RHS is \(y\). To simplify the equation, a common denominator which is a multiple of each of these denominators needs to be found. The denominator \(y^{2}+5 y\) can be factored as \(y*(y+5)\), and therefore this is a common denominator for the fractions in the equation.

Step 2: Transforming the equation using the common denominator

The equation is transformed by doing the cross multiplication. The equation will become: \[ \frac{3 y+5}{y*(y+5)}+\frac{y+4}{y+5}=\frac{y+1}{y} \\ (y+4) * y + (3y+5)*y = (y+1) *(y*(y+5)) \] which simplifies to: \[ y^2+4y + 3y^2+5y = y^2+1y^2+5y \]

Step 3: Simplifying the equation

Simplify the equation by collecting similar terms: \[ y^2+4y + 3y^2+5y - y^2-1y^2-5y = 0 \] which simplifies to: \[4y^2 -2y = 0 \]

Step 4: Solving the equation

You can solve the equation by factoring: \[2y(2y -1) = 0 \] This gives solutions when \(2y = 0\) or \(2y-1 = 0\). So the solutions are \(y = 0\) and \(y = \frac{1}{2}\). Therefore, the solutions to the equation are \(y = 0\) and \(y = \frac{1}{2}\).

Key Concepts

Common Denominator

When you're dealing with fractions in algebraic equations, finding a common denominator is crucial. Think of a common denominator as a shared base for all the fractions involved. This allows us to combine them more easily, just like how finding a common language helps communication. In the problem given, the denominators of the fractions were different:

• The first fraction had a denominator of \( y^2 + 5y \)
• The second fraction had a denominator of \( y + 5 \)
• And the right-hand side (RHS) had a denominator of \( y \)

To find a common denominator, you can start by trying to factorize the expressions. Noticing that \( y^2 + 5y \) can be factored into \( y(y+5) \) helps greatly here. Thus, the common denominator that works is \( y(y+5) \), since it is a multiple of all the denominators in the equation. This transformation allows us to handle the equation uniformly, facilitating the steps that follow.

Factoring

Factoring is a method used to break down algebraic expressions into simpler pieces that are multiplied together. Essentially, it means finding numbers or expressions that multiply together to give the original expression. In the exercise, factoring was used on the denominator \( y^2 + 5y \) to make it \( y(y+5) \).
This step is important because it simplifies the process of finding a common denominator or solving an equation.

• Factoring helps in simplifying expressions by breaking them down
• It leads to easier computation and more straightforward simplifications
• It is particularly useful in solving polynomial equations

When you attempt to solve an equation by setting each factor to zero, you uncover potential solutions to the equation. Such factoring skills are fundamental, helping students streamline their problem-solving process.

Cross Multiplication

Cross multiplication is an effective technique used to eliminate fractions from an equation by multiplying in a particular pattern. When dealing with complex fractions, cross multiplication helps clear the denominators, making it easier to solve for the variable. To apply cross multiplication in the problem:

• The components of each fraction are multiplied diagonally across the equals sign
• Thus, combining: \( (y+4) \times y + (3y+5) \times y = (y+1) \times (y(y+5)) \)

This approach works best when both sides of the equation have a fractional form. It results in the creation of a new equation without fractions, enabling straightforward algebraic simplifications later. Cross multiplication is particularly helpful for balancing equations that contain multiple terms in the numerators and denominators.

Simplification

Simplification is the art of rewriting an equation or expression to make it easier to understand and manage. In algebra, it can involve combining like terms, reducing fractions, or breaking down complex terms for easier handling. It was necessary in this problem to turn the long, complicated expression into a more manageable one.

• First, combine similar terms such as \( y^2 + 3y^2 \) and \( 4y + 5y \)
• This leads from \( y^2+4y + 3y^2+5y - y^2-1y^2-5y = 0 \) to the simplified \( 4y^2 - 2y = 0 \)

Once terms are combined, the equation can be made cleaner and easier to solve by factoring further. Simplification is like tidying up a room – it helps highlight what's essential and aligns everything so that the subsequent steps, like solving for variables, are more effective and less error-prone. It's an essential skill that will serve you well across all areas of mathematics.