Question

1.08) a. set of equations:
b. classification:
1.09) a. set of equations:
b. classification:
a. $y = x + 1$ $2y = 2x + 2$
b. $y = x + 2$ $y = x$
c. $y = -x$ $y = x - 2$
d. $y = 2$ $y = 4$
e. $y = -3x - 3$ $y = -2x - 1$

Explanation:

Step1: Analyze the second graph (1.09)

The second graph has a single line? Wait, no, wait, the first graph (1.08) has two parallel lines? Wait, no, the second graph (1.09) shows a single line? Wait, no, looking at the options, let's check the slopes and intercepts.

Wait, the second graph (1.09) has a line with slope 1 (since it goes through (0,1) and (1,2), etc.), so let's check the options:

Option A: \( y = x + 1 \) and \( 2y = 2x + 2 \). Simplify the second equation: divide by 2, get \( y = x + 1 \). So both equations are the same line (coincident). But the first graph (1.08) has two parallel lines? Wait, no, the second graph (1.09) has one line? Wait, no, the first graph (1.08) has two lines, the second (1.09) has one? Wait, no, maybe I misread.

Wait, 1.09's graph: let's see the line. It passes through (0,1) and has slope 1 (since from (0,1) to (1,2) is up 1, right 1). So the equation is \( y = x + 1 \). Now check the options:

Option A: \( y = x + 1 \) and \( 2y = 2x + 2 \) (which is \( y = x + 1 \)). So these are coincident lines (same line). But the 1.09 graph shows a single line? Wait, no, maybe the 1.09 graph is a single line, meaning the two equations are the same (coincident). Wait, but the options:

Wait, let's re-express each option:

A: \( y = x + 1 \) and \( y = x + 1 \) (since \( 2y = 2x + 2 \) simplifies to \( y = x + 1 \)) → coincident (same line)

B: \( y = x + 2 \) (slope 1, intercept 2) and \( y = x \) (slope 1, intercept 0) → parallel (different intercepts, same slope)

C: \( y = -x \) (slope -1) and \( y = x - 2 \) (slope 1) → intersecting (different slopes)

D: \( y = 2 \) and \( y = 4 \) → horizontal lines, parallel (slope 0, different intercepts)

E: \( y = -3x - 3 \) (slope -3) and \( y = -2x - 1 \) (slope -2) → intersecting (different slopes)

Now, the 1.09 graph: the line has slope 1, intercept 1. So the equation is \( y = x + 1 \). So the set of equations should be A, because both equations in A are \( y = x + 1 \), so they are coincident (same line), which would be represented by a single line (since they overlap).

For classification: coincident lines (same line, infinitely many solutions)

So for 1.09:

a. Set of equations: A. \( y = x + 1 \), \( 2y = 2x + 2 \)

b. Classification: Coincident (or dependent, same line)

Step1: Identify the graph for 1.09

The graph for 1.09 has a single line (since the two equations are the same). Check the options:

Option A: Both equations simplify to \( y = x + 1 \), so they represent the same line (coincident). The graph shows a single line, so this matches.

Step2: Classify the system

Since both equations are the same (coincident), the system is dependent (infinitely many solutions), or coincident lines.

Answer:

a. A. \( y = x + 1 \), \( 2y = 2x + 2 \)
b. Coincident (or Dependent, Same Line)