Question

5.01: proportional relationships
you may use the worksheet or your own lined paper to complete your handwritten work using
the color black.
determine if the values in each table represents a direct variation.
if yes, (a) identify the constant of variation and (b) write an equation to represent the relationship.
no decimals or mixed numbers.
show all work (i.e.. prove)!
1.

x	y
4	12
5	15
6	18

2.

x	y
0	-6
2	-5
4	-4

3.

x	y
-2	1
0	0
4	-2

4.

gallons	miles
2	42
3	63
4	84

5.

text messages	cost ($)
95	7.85
160	9.80
200	11

determine if the equation or graph represents a direct variation. if yes, identify the constant of
variation. no decimals or mixed numbers. show the work on how you made the determination.

1. ( y = 2x - 3 )

yes or no?
constant of variation

1. ( x + y = 9 )

yes or no?
constant of variation

1. ( 4y = x )

yes or no?
constant of variation

Explanation:

Problem 1 (Table 1)

Step1: Recall Direct Variation

Direct variation is \( y = kx \), so \( \frac{y}{x}=k \) (constant).
Check \( \frac{y}{x} \) for each pair:

• \( \frac{9}{3}=3 \)
• \( \frac{12}{4}=3 \)
• \( \frac{15}{5}=3 \)
• \( \frac{18}{6}=3 \)

Step2: Identify Constant and Equation

(a) Constant \( k = 3 \).
(b) Equation: \( y = 3x \).

Step1: Check \( \frac{y}{x} \)

For \( x=-2, y=-7 \): \( \frac{-7}{-2}=\frac{7}{2} \)
For \( x=0, y=-6 \): Division by 0 (undefined, and \( x=0 \) in direct variation requires \( y=0 \), but \( y=-6
eq 0 \)).

Step2: Conclusion

Not direct variation (fails at \( x=0 \) and ratios not constant).

Step1: Check \( \frac{y}{x} \)

For \( x=-6, y=3 \): \( \frac{3}{-6}=-\frac{1}{2} \)
For \( x=-2, y=1 \): \( \frac{1}{-2}=-\frac{1}{2} \)
For \( x=0, y=0 \): \( \frac{0}{0} \) (undefined, but \( x=0,y=0 \) is allowed in direct variation if other ratios constant).
For \( x=4, y=-2 \): \( \frac{-2}{4}=-\frac{1}{2} \)

Step2: Identify Constant and Equation

(a) Constant \( k = -\frac{1}{2} \).
(b) Equation: \( y = -\frac{1}{2}x \).

Answer:

(a) Constant of variation: \( 3 \)
(b) Equation: \( y = 3x \)

Problem 2 (Table 2)