Problem 22 - Algebra A Combined Function

Question

Solve each equation and check each proposed solution. See Examples 4 through 6. $$ \frac{5 y}{y+1}-\frac{3}{y+1}=4 $$

Step-by-Step Solution

Answer

The solution is \( y = 7 \).

Step 1: Understand the Equation

The given equation is \( \frac{5y}{y+1} - \frac{3}{y+1} = 4 \). Notice that both terms on the left side have the same denominator \( y+1 \). This indicates that we can combine them under a single fractional expression.

Step 2: Combine the Fractions

Since both terms \( \frac{5y}{y+1} \) and \( \frac{3}{y+1} \) have the same denominator, we can combine them into a single fraction: \[ \frac{5y - 3}{y+1} = 4 \]

Step 3: Clear the Fraction by Multiplying

To eliminate the fraction, multiply both sides of the equation by \( y+1 \): \[ 5y - 3 = 4(y+1) \]

Step 4: Distribute and Simplify

Expand the right side of the equation by distributing the \( 4 \): \[ 5y - 3 = 4y + 4 \]

Step 5: Solve for y

Move all terms involving \( y \) to one side and constant terms to the other side:\[ 5y - 4y = 4 + 3 \] Simplify this: \[ y = 7 \]

Step 6: Verify the Solution

Substitute \( y = 7 \) back into the original equation to check it's correct:\[ \frac{5(7)}{7+1} - \frac{3}{7+1} = 4 \] This simplifies to:\[ \frac{35}{8} - \frac{3}{8} = \frac{32}{8} = 4 \]The left side equals the right side, confirming \( y = 7 \) is a correct solution.

Key Concepts

Combining Fractions

When we deal with rational equations, one useful strategy is combining fractions. It helps make the equation simpler and easier to solve. Here, both terms on the left side of the equation have the same denominator: \( y+1 \). This means they can be combined into one larger fraction.

• To combine these fractions, simply combine the numerators. In this example, the numerators 5y and -3 come together as \(5y - 3\).
• The denominator remains the same: \( y+1 \).

By writing it as \( \frac{5y-3}{y+1} = 4 \), it simplifies our work and puts the fractions into one neat package. This transformation is like gathering all our coins into one single pocket.

Clearing Fractions

To solve the equation, the next step is clearing fractions. This involves getting rid of the fraction so we can deal with a more straightforward equation. Let's explore how we can do this:

• When you have an equation like \( \frac{5y-3}{y+1} = 4 \), multiplicative techniques can be used to clear the denominator.
• Multiply every term on both sides by \( y+1 \) to cancel out the denominator.

This process helps us move from a fractional equation to an equation that doesn't involve any fractions: \(5y - 3 = 4(y+1)\). Once this step is complete, solving for the variable is much more straightforward.

Solving Linear Equations

Solving linear equations is a crucial skill in algebra. Once the fractions are cleared, we move onto solving the resulting equation. Here is how it is done:

• After clearing fractions, our equation is \(5y - 3 = 4(y+1)\).
• Next, we want to get all terms with the variable on one side and constants on the other. In this scenario, shift the 4y to the left side, and the constant terms to the right.
• Perform any necessary simplifications: \(5y - 4y = 4 + 3\).

Thus, solving the equation leads us to find \( y = 7 \). Linear equations like these have straightforward solutions once we isolate the variable.

Distributive Property

The distributive property is a valuable algebraic tool used here to expand expressions. When you have a term multiplied by a sum, like \(4(y+1)\), the distributive property becomes useful.

• Apply the property by multiplying 4 by each term in the parenthesis. This results in \(4y + 4\).
• This step allows for creating a standard form equation, facilitating easier moves later on when isolating variables.

Expanding equations through distribution helps to break down seemingly complex expressions into simpler components. Therefore, understanding and applying the distributive property is vital for effectively working with equations.