Question

first, fill in the table for each equation. note: each table must be submitted and checked separately.
y = x - 1
x | y
0 | -1
1 | 0
y = -1
x | y
0 | -1
1 | -1
line (with a coordinate grid showing x from -6 to 6 and y from -2 to 6)

Explanation:

To solve the problem of filling the tables and graphing the lines, we analyze each equation:

For the equation \( y = x - 1 \):

We can find more points to complete the table. Let's choose \( x = 2 \):

Step 1: Substitute \( x = 2 \) into \( y = x - 1 \)

\( y = 2 - 1 = 1 \)
So when \( x = 2 \), \( y = 1 \).

For the equation \( y = -1 \):

This is a horizontal line where \( y \) is always -1, regardless of the value of \( x \). So for any \( x \) (e.g., \( x = 2 \)), \( y = -1 \).

Graphing the lines:

• For \( y = x - 1 \), plot the points \((0, -1)\), \((1, 0)\), \((2, 1)\) and draw a straight line through them.
• For \( y = -1 \), plot the points \((0, -1)\), \((1, -1)\), \((2, -1)\) and draw a horizontal straight line through them.

If we were to complete the tables with \( x = 2 \):

Table for \( y = x - 1 \):

\( x \)	\( y \)
1	0
2	1

Table for \( y = -1 \):

\( x \)	\( y \)
1	-1
2	-1

(Note: The problem didn't specify a particular \( x \)-value to use for completion, but the above shows how to extend the tables. The graphs would be a line with slope 1 (for \( y = x - 1 \)) and a horizontal line (for \( y = -1 \)) intersecting at \((0, -1)\).)

Answer:

To solve the problem of filling the tables and graphing the lines, we analyze each equation:

For the equation \( y = x - 1 \):

We can find more points to complete the table. Let's choose \( x = 2 \):

Step 1: Substitute \( x = 2 \) into \( y = x - 1 \)

\( y = 2 - 1 = 1 \)
So when \( x = 2 \), \( y = 1 \).

For the equation \( y = -1 \):

This is a horizontal line where \( y \) is always -1, regardless of the value of \( x \). So for any \( x \) (e.g., \( x = 2 \)), \( y = -1 \).

Graphing the lines:

• For \( y = x - 1 \), plot the points \((0, -1)\), \((1, 0)\), \((2, 1)\) and draw a straight line through them.
• For \( y = -1 \), plot the points \((0, -1)\), \((1, -1)\), \((2, -1)\) and draw a horizontal straight line through them.

If we were to complete the tables with \( x = 2 \):

Table for \( y = x - 1 \):

\( x \)	\( y \)
1	0
2	1

Table for \( y = -1 \):

\( x \)	\( y \)
1	-1
2	-1

(Note: The problem didn't specify a particular \( x \)-value to use for completion, but the above shows how to extend the tables. The graphs would be a line with slope 1 (for \( y = x - 1 \)) and a horizontal line (for \( y = -1 \)) intersecting at \((0, -1)\).)