Question

o y = 2x+3; y = -\frac{1}{3}x + 3
o y = 2x, y = -\frac{1}{3}x
o y = 3x+2, y = 3x - 1
o y = -3x+3, y = x+3

Explanation:

Step1: Find the y - intercept of the lines

The first line passes through the point (0, 3), so its y - intercept \(b = 3\). The second line also has a y - intercept of 3 as it passes through (0, 3).

Step2: Calculate the slope of the first non - horizontal line

Using the slope formula \(m=\frac{y_2 - y_1}{x_2 - x_1}\), for the line passing through (0, 3) and (3, 2), \(m_1=\frac{2 - 3}{3-0}=-\frac{1}{3}\).

Step3: Calculate the slope of the second non - horizontal line

The other line has a steeper positive slope. Let's assume two points on it. If we consider the general form \(y=mx + b\) and the y - intercept \(b = 3\). For a line with a positive slope, if we assume another point \((- 1,1)\) (by observing the graph), then \(m_2=\frac{3 - 1}{0+1}=2\).

Step4: Match with the equations

The equations of the lines should be \(y = 2x+3\) and \(y=-\frac{1}{3}x + 3\)

Answer:

\(y = 2x+3;y=-\frac{1}{3}x + 3\)