Problem 44 - Algebra A Combined Function

Question

Determine whether each pair of lines is parallel, perpendicular, or neither. See Example 7. $$ \begin{array}{l} 10+3 x=5 y \\ 5 x+3 y=1 \end{array} $$

Step-by-Step Solution

Answer

The lines are neither parallel nor perpendicular.

Step 1: Rewrite each equation in slope-intercept form

The slope-intercept form of a line is \( y = mx + b \), where \( m \) is the slope. To convert equations to this form, solve each for \( y \). Starting with the first equation:\[ 10 + 3x = 5y \]Subtract \(3x\) from both sides:\[ 5y = -3x + 10 \]Divide every term by 5:\[ y = -\frac{3}{5}x + 2 \]Now for the second equation:\[ 5x + 3y = 1 \]Subtract \(5x\) from both sides:\[ 3y = -5x + 1 \]Divide every term by 3:\[ y = -\frac{5}{3}x + \frac{1}{3} \]

Step 2: Identify the slopes of each line

From the equations we've converted:1. First line: \( y = -\frac{3}{5}x + 2 \). The slope is \( m_1 = -\frac{3}{5} \).2. Second line: \( y = -\frac{5}{3}x + \frac{1}{3} \). The slope is \( m_2 = -\frac{5}{3} \).

Step 3: Determine if the lines are parallel, perpendicular, or neither

For two lines to be **parallel**, their slopes must be equal. Here, \(-\frac{3}{5} eq -\frac{5}{3}\), so they are not parallel.For two lines to be **perpendicular**, the product of their slopes must equal \(-1\). Calculate the product:\[(-\frac{3}{5}) \times (-\frac{5}{3}) = \frac{15}{15} = 1\]Since the product is 1 (not -1), the lines are not perpendicular.Since the lines are neither parallel nor perpendicular, we conclude they are neither.

Key Concepts

Slope-Intercept Form

The slope-intercept form is one of the most common ways to express a linear equation. It is written as \( y = mx + b \), where \( m \) represents the slope of the line, and \( b \) denotes the y-intercept. This form is particularly convenient for quickly identifying both the slope and the y-intercept, which are key features of the line.

• The slope, \( m \), indicates how steep the line is and the direction it slants. A positive slope implies the line rises as it moves from left to right, while a negative slope means it falls.
• The y-intercept, \( b \), is the point where the line crosses the y-axis. This tells you where the line intersects the vertical axis, aiding in graph plotting and visual understanding.

This format allows for easy comparisons of multiple lines, especially when determining if lines are parallel or perpendicular. Simply put, when two lines are expressed in slope-intercept form, it becomes straightforward to compare their slopes and y-intercepts.

Linear Equations

A linear equation forms a straight line when graphed on a coordinate plane, making it a fundamental concept in algebra. These equations typically take the form \( ax + by = c \) but can be rearranged or simplified into different forms, such as the slope-intercept form. Linear equations are characterized by:

• No exponents higher than one: This keeps the graph a straight line.
• Two-variable relationships: Typically \( x \) and \( y \) are used, representing the horizontal and vertical axes, respectively.

Solving linear equations for a specific variable often involves isolating that variable using algebraic techniques, which aligns with solving for \( y \) in the slope-intercept form. Understanding these equations is crucial for graphing and analysis, as they form the backbone of algebraic thinking and can also be used to model real-world scenarios like predicting future trends.

Algebraic Manipulation

Algebraic manipulation refers to the process of rearranging and simplifying equations to solve for a particular variable. This technique is vital in transforming standard or point-slope equations into the slope-intercept form. Some common steps include:

• Rearranging terms: Moving terms across the equation to isolate variables. For example, adjusting \( ax + by = c \) to \( by = -ax + c \).
• Dividing by coefficients: When isolating \( y \), divide each term by its coefficient if \( y \) isn't alone. This is crucial in achieving the simple \( y = mx + b \) form.

Through algebraic manipulation, one can easily identify the slope and y-intercept, making it simpler to analyze the line's behavior and relationships with other lines. Mastering these skills lays a strong foundation for tackling more complex algebraic problems.