1. Direct Variation
Lesson 2-3

2. = and are both equal to –3, but = 3.
Direct Variation
Lesson 2-3
Additional Examples
For each function, determine whether y varies directly with x. If so, find the constant of variation and write the equation.
x –
y 3 –6 15 a.
= and are both equal to –3, but = 3. y x 3 –1 –6 2 15 5
Since the three ratios are not all equal, y does not vary directly with x.
x –4
y –8 b.
= 2, so y does vary directly with x. y x 14 7 18 9 –8 –4 =
The constant of variation is 2.
The equation is y = 2x.

3. Determine whether y varies directly with x. If so, write the equation.

4. 5x = –2y is equivalent to y = – x, so y varies directly with x.
Direct Variation
Lesson 2-3
Additional Examples
For each function, tell whether y varies directly with x. If so, find the constant of variation.
a. 3y = 7x + 7
Since you cannot write the equation in the form y = kx, y does not vary directly with x.
b. 5x = –2y
5x = –2y is equivalent to y = – x, so y varies directly with x. 5 2
The constant of variation is – . 5 2

5. a. Find the constant of variation.
Direct Variation
Lesson 2-3
Additional Examples
The perimeter of a square varies directly as the length of a side of the square. The formula P = 4s relates the perimeter to the length of a side.
a. Find the constant of variation.
The equation P = 4s has the form of a direct variation equation with k = 4.
b. Find how long a side of the square must be for the perimeter to be 64 cm.
P = 4s Use the direct variation.
64 = 4s Substitute 64 for P.
16 = s Solve for s.
The sides of the square must have length 16 cm.

6. 15(18) = 27(y) Write the cross products.
Direct Variation
Lesson 2-3
Additional Examples
Suppose y varies directly with x, and y = 15 when x = 27. Find y when x = 18. Let (x1, y1) = (27, 15) and let (x2, y2) = (18, y).
Write a proportion. y1 x1 y2 x2 =
Substitute. = 15 27 y 18
15(18) = 27(y) Write the cross products.
y = Solve for y.
15 • 18 27
y = 10 Simplify.

7. Find the missing value for the direct variation: If y = 4 when x = 3, what is y when x = 6?