Problem 45 - Algebra for College Students

Question

$$ (y+1)(a-3)=x-2 \text { for } y $$

Step-by-Step Solution

Answer

\( y = \frac{x + 1 - a}{a - 3} \)

Step 1: Distribute Terms

To isolate \( y \), begin by expanding the left-hand side of the equation using the distributive property. Multiply each term within the parenthesis: \( (y+1)(a-3) = ya - 3y + a - 3 \). The equation now reads \( ya - 3y + a - 3 = x - 2 \).

Step 2: Rearrange Equation for y

Move all terms that do not involve \( y \) to the right-hand side of the equation to isolate \( y \). We have \( ya - 3y = x - 2 - a + 3 \). Simplify the right-hand side to get \( x + 1 - a \).

Step 3: Factor Out y

Factor \( y \) out from the left-hand side: \( y(a - 3) = x + 1 - a \). This makes \( y \) easier to isolate.

Step 4: Solve for y

Divide both sides by \( a - 3 \) to solve for \( y \): \( y = \frac{x + 1 - a}{a - 3} \). This fraction gives the value of \( y \) in terms of \( x \) and \( a \).

Key Concepts

Distributive Property

The distributive property is a fundamental concept in algebra that helps simplify expressions and solve equations. It involves distributing a factor across terms inside a parenthesis. This is done by multiplying the outside term with each term inside the parentheses.
For example, in the equation \( (y+1)(a-3) \), applying the distributive property involves multiplying \( (y+1) \) by each term inside \((a-3)\):

• \( y \times a = ya \)
• \( y \times -3 = -3y \)
• \( 1 \times a = a \)
• \( 1 \times -3 = -3 \)

This results in the expanded expression \( ya - 3y + a - 3 \). By using the distributive property correctly, you prepare the equation for further steps like isolating variables.

Factoring Equations

Factoring is the process of rewriting an expression or equation by extracting common factors, making it easier to solve for variables. It's particularly useful when you aim to simplify polynomial expressions or look for solutions in algebra.
In the original exercise, after using the distributive property to expand the equation, you end up with terms like \( ya \) and \( -3y \) on the left-hand side. Since y is common in both terms, you can factor it out:

• Start with \( ya - 3y = y(a - 3) \)

Factoring here simplifies the equation significantly, grouping all terms with \( y \) together, which brings you a step closer to isolating y. It emphasizes the part where \( y \) is involved, making further steps for solving the equation much clearer.

Isolating Variables

Isolating a variable means rearranging an equation so that a specific variable stands alone on one side of the equation, ideally solving it in terms of other variables. This concept is crucial as it simplifies understanding the relationships between variables.
In the exercise, once factoring is done, the equation \( y(a - 3) = x + 1 - a \) is left. To isolate \( y \), the next step is to make \( y \) the sole subject of the equation:

• Divide both sides by \( (a - 3) \),
• This results in \( y = \frac{x + 1 - a}{a - 3} \)

By isolating \( y \), you determine its value strictly in terms of \( x \) and \( a \). Isolating variables is a common final step in solving algebraic equations, providing a clear and concise solution.