Problem 15 - Algebra A Combined Function

Question

Solve each equation and check each proposed solution. See Examples 4 through 6. $$ \frac{4 y}{y-4}+5=\frac{5 y}{y-4} $$

Step-by-Step Solution

Answer

The solution is \(y = 5\). Verified by substituting back, both sides equal 25.

Step 1: Understand the Equation

The given equation is \(\frac{4y}{y-4} + 5 = \frac{5y}{y-4}\). It is a rational equation because it contains fractions with a variable in the denominator.

Step 2: Eliminate the Denominator

To eliminate the denominator, multiply every term by \(y-4\), the common denominator: \((y-4)\left(\frac{4y}{y-4} + 5\right) = (y-4)\frac{5y}{y-4}\).This simplifies to: \[4y + 5(y-4) = 5y\].

Step 3: Simplify the Equation

Distribute and simplify the terms: \[4y + 5y - 20 = 5y\].Combine like terms: \[9y - 20 = 5y\].

Step 4: Solve for y

Move all the terms with \(y\) to one side of the equation: \[9y - 5y = 20\].Simplifying this gives: \[4y = 20\].Finally, divide both sides by 4 to solve for \(y\): \[y = 5\].

Step 5: Verify the Solution

Plug \(y = 5\) back into the original equation to verify:\(\frac{4(5)}{5-4} + 5 = \frac{5(5)}{5-4}\).Both sides simplify to 25, so \(y = 5\) is indeed a solution.

Key Concepts

Rational Equations

Understanding rational equations is essential because they frequently appear in algebra. A rational equation is basically an equation that involves one or more fractions which have variables in the denominator.
These equations are an extension of rational expressions. The term "rational" means that the expressions can be written as fractions (or ratios), where both numerator and denominator are polynomials.
When you see a rational equation like \( \frac{4y}{y-4} + 5 = \frac{5y}{y-4} \), the goal is to solve for the variable, which in this case is \( y \). Solving a rational equation often involves finding a common denominator to eliminate the fractions and simplify the problem.

• Keep an eye out for restrictive solutions. These occur when the value of the variable makes any denominator zero.
• Identify the least common denominator (LCD) among the fractions to eliminate the denominators effectively.

This process will transform the equation into a simpler form, making it easier to solve.

Equation Verification

After solving a rational equation, it is crucial to verify the proposed solution to ensure accuracy. Verification helps confirm that no mistakes were made during simplification or solving.
Verification involves plugging the solution back into the original equation to check if both sides of the equation are equal. If they are equal, it confirms the solution's correctness.
Let's consider the earlier solution \( y = 5 \). To verify if this is correct, substitute \( y = 5 \) back into the original equation:

• Substitute: \( \frac{4(5)}{5-4} + 5 = \frac{5(5)}{5-4} \)
• Simplify both sides: \( 20/1 + 5 = 25/1 \)

Since both sides equal 25, \( y = 5 \) is indeed the correct solution.
Remember, verifying is not just a "safety-check," it's an essential step in ensuring the reliability of your solution.

Simplifying Equations

Simplifying equations is a foundational skill needed to solve rational equations efficiently. The process involves refining the equation to its simplest form using algebraic rules.
Starting with an equation like \( 4y + 5(y-4) = 5y \), we simplify by performing operations that condense terms into fewer, simpler ones.
Here's how simplification works:

• Distribute any constants or coefficients through parentheses: e.g., \( 5(y-4) \) becomes \( 5y - 20 \).
• Combine like terms: Combine terms that contain the same variable, here \( 4y + 5y = 9y \).
• Solve for the variable: Rearrange the equation to isolate the variable on one side.

In our example, simplifying gave us \( 9y - 20 = 5y \) which was further solved to \( 4y = 20 \). Simplifying isn't just about making an equation smaller; it's about making it easier to handle and solve, paving the way for an accurate solution.