91Ó°ÊÓ

In a direct variation, the relationship between two variables, typically represented as \( x \) and \( y \), is such that one variable is a constant multiple of the other. This constant multiplier is known as the 'constant of proportionality' and is usually denoted by \( k \). The general formula for direct variation is \( y = kx \). In this problem, we were given \( x = 1 \) and \( y = 5 \) and required to find \( k \).
By substituting these values into the equation, \( 5 = k \times 1 \), we easily solve for \( k \). Therefore, \( k = 5 \). This constant value helps us understand how the changes in \( x \) affect \( y \).
Remember, the constant of proportionality can also be referred to as the 'slope' in some contexts within linear relationships.

The equation derived from direct variation can be written in the 'slope-intercept form', which is generally expressed as \( y = mx + b \). In this equation, \( m \) represents the slope, and \( b \) represents the y-intercept. For direct variation problems, the y-intercept \( b \) is always zero because when \( x = 0 \), \( y \) also equals 0.
So, the slope-intercept form of the equation in this case is \( y = 5x + 0 \), which simplifies to \( y = 5x \). Here, the slope \( m = 5 \) indicates how steep the line is on a graph. The larger the slope, the steeper the line. Since the y-intercept is 0, the line passes through the origin (0,0).

Once we have the equation in the slope-intercept form, it becomes straightforward to graph it. The equation \( y = 5x \) can be graphed by following these steps:
1. Plot the y-intercept (0,0) on the graph.
2. Use the slope to find another point. Since the slope is 5, starting from (0,0), move 5 units up and 1 unit to the right to locate the point (1,5).
3. Draw a straight line through these points.
The line you draw represents all the possible solutions to the equation \( y = 5x \). Any point on this line is a valid (\( x, y \)) pair that satisfies the equation.

Algebraic substitution is a method used to reveal unknown values by replacing variables with given numbers. In our direct variation problem, substitution was significant in multiple steps:
- To find the constant of proportionality, \( k \), we substituted \( x = 1 \) and \( y = 5 \) into the equation \( y = kx \).
- To determine \( y \) when \( x = 2 \), we substituted \( x = 2 \) into the equation \( y = 5x \), yielding \( y = 10 \).
- Finally, to find \( y \) when \( x = 3 \), we substituted \( x = 3 \) into the same equation, resulting in \( y = 15 \).
Substitution simplifies finding specific values and confirming results by plugging in known quantities into your equations.