Question

module 11 (linear equations) assignment
score: 7/20 answered: 7/20
question 8
sketch a graph of the linear equation x = 3y.
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Explanation:

Step1: Rewrite the equation in slope - intercept form

The given equation is \(x = 3y\). We can rewrite it as \(y=\frac{1}{3}x\). The slope - intercept form of a linear equation is \(y = mx + b\), where \(m\) is the slope and \(b\) is the y - intercept. In this case, \(m=\frac{1}{3}\) and \(b = 0\).

Step2: Find two points on the line

• When \(x = 0\), substitute into \(y=\frac{1}{3}x\), we get \(y = 0\). So the point \((0,0)\) is on the line.
• When \(x=3\), substitute into \(y = \frac{1}{3}x\), we have \(y=\frac{1}{3}\times3 = 1\). So the point \((3,1)\) is on the line.

Step3: Plot the points and draw the line

Plot the points \((0,0)\) and \((3,1)\) on the coordinate plane. Then draw a straight line passing through these two points. The line should have a positive slope (since \(m=\frac{1}{3}>0\)) and pass through the origin.

(Note: Since the problem asks to sketch the graph, the key is to identify the slope and the y - intercept, find two points, and draw the line. The above steps describe how to determine the points and the nature of the line for sketching.)

Answer:

To sketch the graph of \(x = 3y\) (or \(y=\frac{1}{3}x\)):

1. Identify that the line passes through the origin \((0,0)\) (since \(b = 0\)) and has a slope of \(\frac{1}{3}\) (for every 3 units you move to the right along the x - axis, you move up 1 unit along the y - axis).
2. Plot the points \((0,0)\) and \((3,1)\) (or other points like \((- 3,-1)\)) on the coordinate grid.
3. Draw a straight line passing through these points.