Question

4/20
use the given equation to complete the
table.
( y = \frac{2}{3}x - 1 )
drag labels to their correct position on the image
= 7
= 1
= 5
= -3
= -5
= -7
= -1

x
6
-3
0
3

y

Explanation:

Step1: For \( x = 6 \)

Substitute \( x = 6 \) into \( y=\frac{2}{3}x - 1 \).
\( y=\frac{2}{3}\times6 - 1 = 4 - 1 = 3 \)

Step2: For \( x = - 3 \)

Substitute \( x=-3 \) into \( y=\frac{2}{3}x - 1 \).
\( y=\frac{2}{3}\times(-3)-1=-2 - 1=-3 \)

Step3: For \( x = 0 \)

Substitute \( x = 0 \) into \( y=\frac{2}{3}x - 1 \).
\( y=\frac{2}{3}\times0 - 1=0 - 1=-1 \)

Step4: For \( x = 3 \)

Substitute \( x = 3 \) into \( y=\frac{2}{3}x - 1 \).
\( y=\frac{2}{3}\times3 - 1=2 - 1 = 1 \)
Wait, there may be a mis - reading of the \( x \) values. Let's re - check the table. If the \( x \) values are \( 6,-3,0,3 \) (assuming the third \( x \) value is \( - 3 \) as per the hand - like icon covering maybe a negative sign). Wait, maybe the \( x \) values are \( 6, - 6,0,3 \)? Let's re - calculate with \( x=-6 \):
Substitute \( x = - 6 \) into \( y=\frac{2}{3}x-1 \), \( y=\frac{2}{3}\times(-6)-1=-4 - 1=-5 \)
Substitute \( x = 6 \): \( y=\frac{2}{3}\times6-1 = 4 - 1 = 3 \)
Substitute \( x=-3 \): \( y=\frac{2}{3}\times(-3)-1=-2 - 1=-3 \)
Substitute \( x = 0 \): \( y=\frac{2}{3}\times0-1=-1 \)
Substitute \( x = 3 \): \( y=\frac{2}{3}\times3-1=2 - 1 = 1 \)

If we assume the \( x \) values are \( 6,-6,0,3 \) (maybe a typo in the original table display), then:

• When \( x = 6 \), \( y = 3 \)
• When \( x=-6 \), \( y=\frac{2}{3}\times(-6)-1=-4 - 1=-5 \)
• When \( x = 0 \), \( y=-1 \)
• When \( x = 3 \), \( y = 1 \)

But from the given number labels \( 7,-7,1,5,-3,-1,-9 \), maybe the correct \( x \) values and calculations are:

Let's list the \( x \) values as per the table (from left to right, the \( x \) column has \( x \) values: let's assume the \( x \) values are \( 6,-3,0,3 \) (the order of the table rows: first \( x = 6 \), second \( x=-3 \), third \( x = 0 \), fourth \( x = 3 \))

1. For \( x = 6 \):

\( y=\frac{2}{3}\times6-1=4 - 1 = 3 \)

1. For \( x=-3 \):

\( y=\frac{2}{3}\times(-3)-1=-2 - 1=-3 \)

1. For \( x = 0 \):

\( y=\frac{2}{3}\times0-1=-1 \)

1. For \( x = 3 \):

\( y=\frac{2}{3}\times3-1=2 - 1 = 1 \)

Answer:

When \( x = 6 \), \( y = 3 \); when \( x=-3 \), \( y=-3 \); when \( x = 0 \), \( y=-1 \); when \( x = 3 \), \( y = 1 \)

If we consider the number labels, and assume the \( x \) values are \( 9,-6,0,3 \) (maybe a mis - print in the original problem), let's recalculate:

• \( x = 9 \): \( y=\frac{2}{3}\times9-1=6 - 1 = 5 \)
• \( x=-6 \): \( y=\frac{2}{3}\times(-6)-1=-4 - 1=-5 \)
• \( x = 0 \): \( y=-1 \)
• \( x = 3 \): \( y = 1 \)

Given the labels \( 7,-7,1,5,-3,-1,-9 \), the most probable correct calculations are:

For \( x = 6 \): \( y = 3 \) (but 3 is not in the labels? Wait, maybe the equation is \( y=\frac{2}{3}x + 1 \)? Let's check:

If \( y=\frac{2}{3}x + 1 \):

• \( x = 6 \): \( y=\frac{2}{3}\times6+1=4 + 1 = 5 \)
• \( x=-6 \): \( y=\frac{2}{3}\times(-6)+1=-4 + 1=-3 \)
• \( x = 0 \): \( y=0 + 1 = 1 \)
• \( x = 3 \): \( y=\frac{2}{3}\times3+1=2 + 1 = 3 \) (still 3 not in labels)

Wait, maybe the equation is \( y=\frac{2}{3}x-1 \) and the \( x \) values are \( 12,-9,0,6 \):

• \( x = 12 \): \( y=\frac{2}{3}\times12-1=8 - 1 = 7 \)
• \( x=-9 \): \( y=\frac{2}{3}\times(-9)-1=-6 - 1=-7 \)
• \( x = 0 \): \( y=-1 \)
• \( x = 6 \): \( y = 3 \)

Ah! This matches the labels \( 7,-7,1,5,-3,-1,-9 \). So if \( x = 12 \), \( y = 7 \); \( x=-9 \), \( y=-7 \); \( x = 0 \), \( y=-1 \); \( x = 6 \), \( y = 3 \); \( x=-6 \), \( y=-5 \); \( x = 3 \), \( y = 1 \)

Let's do the correct calculation with the assumption that the \( x \) values are \( 12,-9,0,6, - 6,3 \) (to match the number of labels):

1. \( x = 12 \):

\( y=\frac{2}{3}\times12-1=8 - 1 = 7 \)

1. \( x=-9 \):

\( y=\frac{2}{3}\times(-9)-1=-6 - 1=-7 \)

1. \( x = 0 \):

\( y=\frac{2}{3}\times0-1=-1 \)

1. \( x = 6 \):

\( y=\frac{2}{3}\times6-1=4 - 1 = 3 \)

1. \( x=-6 \):

\( y=\frac{2}{3}\times(-6)-1=-4 - 1=-5 \)

1. \( x = 3 \):

\( y=\frac{2}{3}\times3-1=2 - 1 = 1 \)

The correct pairings (assuming the table has \( x \) values \( 12,-9,0,6,-6,3 \)) are:

• \( x = 12 \) corresponds to \( y = 7 \)
• \( x=-9 \) corresponds to \( y=-7 \)
• \( x = 0 \) corresponds to \( y=-1 \)
• \( x = 6 \) corresponds to \( y = 3 \)
• \( x=-6 \) corresponds to \( y=-5 \)
• \( x = 3 \) corresponds to \( y = 1 \)

If we take the original table as having \( x \) values \( 6,-6,0,3 \) (four values) and the labels as \( 3,-5,-1,1 \) (matching our calculation when \( x = 6,y = 3 \); \( x=-6,y=-5 \); \( x = 0,y=-1 \); \( x = 3,y = 1 \))

So the completed table (for \( x = 6,-6,0,3 \)):

\( x \)	\( 6 \)	\( - 6 \)	\( 0 \)	\( 3 \)

And if we consider the extra labels \( 7,-7,-9 \), maybe there are more \( x \) values like \( 12,-9,0 \) (for \( x = 12,y = 7 \); \( x=-9,y=-7 \); \( x = 0,y=-1 \))