Question

tell whether x and y show direct variation. explain your reasoning. if so, find k.
6.

x	1	2	3	4
y	2	4	6	8

7.

x	-2	-1	0	1
y	0	2	4	6

8.

x	-1	0	1	2
y	-2	-1	0	1

9.

x	3	6	9	12
y	2	4	6	8

1. (partially visible)

Explanation:

Step1: Recall Direct Variation Definition

Direct variation is a relationship between two variables \( x \) and \( y \) such that \( y = kx \), where \( k \) is a constant (the constant of variation). This means the ratio \( \frac{y}{x} \) should be constant for all non - zero \( x \) values, and when \( x = 0 \), \( y \) must also be \( 0 \) (since \( y=k\times0 = 0 \)).

Step2: Analyze Problem 6

For the table in problem 6:

• When \( x = 1 \), \( y = 2 \), so \( \frac{y}{x}=\frac{2}{1}=2 \)
• When \( x = 2 \), \( y = 4 \), so \( \frac{y}{x}=\frac{4}{2}=2 \)
• When \( x = 3 \), \( y = 6 \), so \( \frac{y}{x}=\frac{6}{3}=2 \)
• When \( x = 4 \), \( y = 8 \), so \( \frac{y}{x}=\frac{8}{4}=2 \)

The ratio \( \frac{y}{x} \) is constant (\( k = 2 \)) for all \( x\) values, and if we consider the direct variation formula \( y=kx \), when \( x = 0 \) (not in the table, but by the formula), \( y = 0 \). So \( x \) and \( y \) show direct variation with \( k = 2 \).

Step3: Analyze Problem 7

For the table in problem 7:

• When \( x=- 2\), \( y = 0 \), \( \frac{y}{x}=\frac{0}{-2}=0 \)
• When \( x=-1\), \( y = 2 \), \( \frac{y}{x}=\frac{2}{-1}=-2 \)
• The ratios are not constant, and also when \( x = 0 \), \( y = 4

eq0 \). So \( x \) and \( y \) do not show direct variation.

Step4: Analyze Problem 8

For the table in problem 8:

• When \( x=-1\), \( y=-2 \), \( \frac{y}{x}=\frac{-2}{-1}=2 \)
• When \( x = 0 \), \( y=-1

eq0 \) (violates the direct variation condition that \( y = 0 \) when \( x = 0 \) for \( y=kx \)). So \( x \) and \( y \) do not show direct variation.

Step5: Analyze Problem 9

For the table in problem 9:

• When \( x = 3 \), \( y = 2 \), \( \frac{y}{x}=\frac{2}{3}\approx0.666\)
• When \( x = 6 \), \( y = 4 \), \( \frac{y}{x}=\frac{4}{6}=\frac{2}{3}\approx0.666\)
• When \( x = 9 \), \( y = 6 \), \( \frac{y}{x}=\frac{6}{9}=\frac{2}{3}\approx0.666\)
• When \( x = 12 \), \( y = 8 \), \( \frac{y}{x}=\frac{8}{12}=\frac{2}{3}\approx0.666\)

The ratio \( \frac{y}{x} \) is constant (\( k=\frac{2}{3} \)) for all \( x \) values. So \( x \) and \( y \) show direct variation with \( k=\frac{2}{3} \).

Answer:

• Problem 6: \( x \) and \( y \) show direct variation. The constant of variation \( k = 2 \) (since \( \frac{y}{x}=2 \) for all \( x \) values).
• Problem 7: \( x \) and \( y \) do not show direct variation (ratios are not constant and \( y

eq0 \) when \( x = 0 \)).

• Problem 8: \( x \) and \( y \) do not show direct variation (\( y

eq0 \) when \( x = 0 \) and ratios are not consistent in a way that fits \( y = kx \)).

• Problem 9: \( x \) and \( y \) show direct variation. The constant of variation \( k=\frac{2}{3} \) (since \( \frac{y}{x}=\frac{2}{3} \) for all \( x \) values).