Question

problem 7
match each equation with the graph of its line.
equations: ( y = \frac{1}{3}x ), ( y = \frac{2}{3}x + 2 ), ( y = \frac{3}{2}x + 1 )
graphs: three coordinate plane graphs with lines labeled r, q, p (grid with x and y axes from 0 to 10)

Explanation:

Step1: Analyze \( y = \frac{1}{3}x \)

This is a linear equation in slope - intercept form \( y=mx + b \) where \( m=\frac{1}{3} \) (slope) and \( b = 0 \) (y - intercept). So the line passes through the origin \((0,0)\) and has a positive slope. Let's check the graphs. The bottom - most graph (with point P) seems to pass through the origin. Let's verify the slope. For \( y=\frac{1}{3}x \), when \( x = 3 \), \( y = 1 \); when \( x=6 \), \( y = 2 \), etc. The bottom graph (with P) has a slope that looks like \( \frac{1}{3} \) (rise over run: for example, from \((0,0)\) to \((3,1)\), rise = 1, run = 3, slope=\( \frac{1}{3} \)). So \( y=\frac{1}{3}x \) matches the bottom graph (with P).

Step2: Analyze \( y=\frac{2}{3}x + 2 \)

The y - intercept \( b = 2 \), so the line crosses the y - axis at \((0,2)\). The slope \( m=\frac{2}{3} \). Let's look at the middle graph (with Q). The line in the middle graph has a y - intercept. Let's check the slope. If we take two points, say, when \( x = 0 \), \( y = 2 \) (matches the y - intercept), and when \( x = 3 \), \( y=\frac{2}{3}(3)+2=2 + 2=4 \)? Wait, no, maybe I made a mistake. Wait, the middle graph (with Q) seems to have a negative slope? Wait, no, maybe I misread. Wait, the top graph (with R) has a y - intercept at \( y = 1 \)? Wait, no, let's re - examine. Wait, the equation \( y=\frac{2}{3}x + 2 \): y - intercept at \( (0,2) \). The middle graph (with Q) has a y - intercept? Wait, maybe I mixed up. Wait, the top graph (with R): let's check its y - intercept. The top graph (with R) starts at \( (0,1) \)? No, wait the first (top) graph: when \( x = 0 \), \( y = 1 \)? Wait, no, the equation \( y=\frac{3}{2}x+1 \): y - intercept at \( (0,1) \), slope \( \frac{3}{2} \). The top graph (with R): let's take two points. If \( x = 0 \), \( y = 1 \), and when \( x = 2 \), \( y=\frac{3}{2}(2)+1=3 + 1 = 4 \)? Wait, no, the top graph (with R) has a steeper slope. Wait, let's re - categorize:

• \( y=\frac{1}{3}x \): passes through (0,0), slope \( \frac{1}{3} \) → bottom graph (P)
• \( y=\frac{3}{2}x + 1 \): y - intercept at (0,1), slope \( \frac{3}{2} \) (steep positive slope). The top graph (with R): when \( x = 0 \), \( y = 1 \), and when \( x = 2 \), \( y=\frac{3}{2}(2)+1=4 \)? Wait, the top graph (with R) has a line that goes from (0,1) to (8, let's see, if \( x = 8 \), \( y=\frac{3}{2}(8)+1=13 \)? No, maybe the top graph (with R) is \( y=\frac{3}{2}x + 1 \)? Wait, no, the equation \( y=\frac{3}{2}x+1 \): slope \( \frac{3}{2} \), y - intercept 1. The top graph (with R) has a y - intercept at \( y = 1 \) (starts at (0,1)) and a steep slope. So \( y=\frac{3}{2}x + 1 \) matches the top graph (with R).

• \( y=\frac{2}{3}x+2 \): y - intercept at (0,2), slope \( \frac{2}{3} \). The middle graph (with Q): let's check the y - intercept. If the middle graph (with Q) has a y - intercept at (0,2), and slope \( \frac{2}{3} \). Let's take \( x = 3 \), \( y=\frac{2}{3}(3)+2=2 + 2 = 4 \), \( x = 6 \), \( y=\frac{2}{3}(6)+2=4 + 2 = 6 \). The middle graph (with Q) seems to have a line that goes through (0,2) and (3,4), etc. So \( y=\frac{2}{3}x + 2 \) matches the middle graph (with Q).

Answer:

• \( y=\frac{1}{3}x \) matches the bottom graph (with point P)
• \( y=\frac{2}{3}x + 2 \) matches the middle graph (with point Q)
• \( y=\frac{3}{2}x + 1 \) matches the top graph (with point R)