Problem 39 - Algebra for College Students

Question

For Problems \(31-44\), solve each equation for the indicated variable. $$ \frac{y-1}{x-3}=\frac{b-1}{a-3} \text { for } y $$

Step-by-Step Solution

Answer

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

Step 1: Cross-Multiply

To eliminate the fractions, cross-multiply the terms. This means multiplying the numerator of the first fraction by the denominator of the second, and vice versa. You get: \[ (y-1)(a-3) = (b-1)(x-3) \]

Step 2: Distribute the Terms

Distribute each binomial multiplication on both sides of the equation. This will eliminate the parentheses:\[ ya - 3y - a + 3 = bx - 3b - x + 3 \]

Step 3: Isolate the Terms Involving y

Move all the terms involving \(y\) to one side of the equation and the constant terms to the other. You get:\[ ya - 3y = bx - 3b - x + a - 3 \]

Step 4: Factor Out the Variable y

Factor \(y\) from the terms on the left-hand side:\[ y(a - 3) = bx - 3b - x + a - 3 \]

Step 5: Solve for y

To solve for \(y\), divide both sides by the coefficient \((a - 3)\):\[ y = \frac{bx - 3b - x + a - 3}{a - 3} \]

Key Concepts

Cross-Multiplication

Cross-multiplication is a technique used to simplify equations that involve fractions. By rearranging the terms, it helps eliminate the fractions, making the equation easier to solve. Here's how it works:

• Identify the two sides of the equation that involve fractions.
• Multiply the numerator of the first fraction by the denominator of the second fraction.
• Do the same for the numerator of the second fraction and the denominator of the first fraction.

This operation produces an equation without fractions. For example, from the exercise, when you cross-multiply \( \frac{y-1}{x-3} = \frac{b-1}{a-3} \), you get \( (y-1)(a-3) = (b-1)(x-3) \). This step is crucial as it clears the fractions early on, simplifying subsequent calculations.

Distributive Property

Once the fractions are eliminated through cross-multiplication, the distributive property comes into play. This property is a fundamental arithmetic rule used to multiply a single term by two or more terms inside a set of parentheses. Here's how it applies:

• Take each term inside the parentheses of one binomial and multiply it by every term inside the parentheses of the connected binomial.
• This removes the parentheses and results in a more straightforward expression.

In the provided solution, distributing terms results in \( ya - 3y - a + 3 = bx - 3b - x + 3 \). Breaking down complex expressions like this is vital for managing and simplifying algebraic equations.

Variable Isolation

The next step after using the distributive property is to isolate the variable you are solving for. This process involves arranging the equation so that the desired variable is alone on one side. Follow these general steps:

• Move all terms involving the target variable \( y \) to one side of the equation.
• Shift all other terms to the opposite side.
• Simplify the expression as much as possible.

In this scenario, moving terms gives: \( ya - 3y = bx - 3b - x + a - 3 \). Isolating the variable lays the groundwork for solving the equation and helps maintain focus on what needs to be calculated.

Factoring

The final maneuver before directly solving for the variable is factoring. Factoring reorganizes terms to simplify the division needed to solve for an isolated variable. Here’s the strategy:

• Identify a common factor in the terms on one side of the equation containing the target variable.
• Factor out that common element, simplifying the expression.
• This typically sets up the final division needed to solve the equation.

In this case, factoring gives \( y(a - 3) = bx - 3b - x + a - 3 \). By factoring out \( y \), it makes the equation ready to divide by the remaining coefficient \( (a-3) \), resulting in \( y = \frac{bx - 3b - x + a - 3}{a - 3} \). This step is pivotal for isolating and solving for the desired variable.