Problem 43 - Algebra for College Students

Question

$$ \frac{y-a}{b}=\frac{x+b}{c} \text { for } x $$

Step-by-Step Solution

Answer

The solution for \( x \) is \( x = \frac{cy - ca - b^2}{b} \).

Step 1: Set up the Equation

We start with the given equation: \( \frac{y-a}{b} = \frac{x+b}{c} \). Our goal is to solve for \( x \).

Step 2: Cross Multiply

To eliminate the fractions, cross multiply: \( (y-a) \cdot c = b \cdot (x+b) \). This simplifies to \( c(y-a) = b(x+b) \).

Step 3: Distribute the Terms

Distribute the terms on both sides of the equation: \( cy - ca = bx + b^2 \).

Step 4: Isolate Terms with x on One Side

Move all terms involving \( x \) to one side and constant terms to the other side: \( bx = cy - ca - b^2 \).

Step 5: Solve for x

Finally, solve for \( x \) by dividing the entire equation by \( b \): \( x = \frac{cy - ca - b^2}{b} \).

Key Concepts

Cross Multiplication

Cross multiplication is a simple technique used to eliminate fractions from equations by multiplying across the equal sign. It is particularly useful in solving equations involving two fractions. In the provided exercise, cross multiplication helps to clear the denominators from both sides of the equation, making it easier to solve for the variable.
To apply cross multiplication, multiply the numerator of one fraction by the denominator of the other fraction, and do the same for the opposite pair. This leads to a new equation that no longer contains fractions.
For example, using the equation \( \frac{y-a}{b} = \frac{x+b}{c} \):

• Multiply \((y-a)\) by \(c\), which gives \((y-a)\cdot c\).
• Multiply \(b\) by \((x+b)\), which results in \(b(x+b)\).
• After cross multiplying, the equation becomes \(c(y-a) = b(x+b)\).

Cross multiplication simplifies the equation to a linear form, making it ready for further algebraic manipulation like distribution or isolating variables.

Distributive Property

The distributive property is often used in algebra to expand expressions that have terms inside parentheses. It allows you to multiply a single term outside the parentheses by each term inside of it, which is essential in simplifying equations.
In the exercise provided, after cross multiplication gives \(c(y-a) = b(x+b)\), the distributive property is applied to both sides of the equation.

• For the left side, distribute \(c\) over \(y-a\), resulting in \(cy - ca\).
• On the right side, distribute \(b\) over \(x+b\), giving \(bx + b^2\).

The equation now reads \(cy - ca = bx + b^2\). This step is crucial as it eliminates the parentheses, allowing terms to be moved and combined, setting the stage to solve for the desired variable.

Isolating Variables

Isolating variables is a core algebraic process used to solve for a particular variable in an equation. The goal is to rearrange the equation so that the variable you are solving for is on one side and everything else is on the other.
In this exercise, after applying the distributive property, the equation becomes \(cy - ca = bx + b^2\). To solve for \(x\), we need to isolate it.

• Move every term not involving \(x\) to the other side: \(bx = cy - ca - b^2\).
• This involves adding or subtracting terms to both sides of the equation as needed.
• Finally, divide each side by the coefficient of \(x\) to solve for \(x\). Here, divide by \(b\) which results in \(x = \frac{cy - ca - b^2}{b}\).

By isolating \(x\), we have successfully solved the equation for the desired variable. Keeping the steps structured ensures clarity and accuracy when performing algebraic manipulations.