Worksheet on direct and inverse variation with practice problems and solutions.

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Section 1: Direct and Inverse Variation Graphs

Problem 1: "Which graph represents a direct variation? Write the equation of the line."
* Analysis: A direct variation is a relationship of the form $y = kx$. This means the graph must be a straight line passing through the origin $(0,0)$.
* Graph A: The line passes through the origin $(0,0)$ and the point $(2, 4)$.
* Graph B: The line does not pass through the origin; it passes through $(0, 2)$.
* Conclusion: Graph A represents direct variation.
* Finding the equation: Using the point $(2, 4)$:
$$y = kx$$
$$4 = k(2)$$
$$k = 2$$
So, the equation is $y = 2x$.

Problem 2: "Which graph represents an inverse variation? Write the equation of the curve."
* Analysis: An inverse variation is a relationship of the form $y = k/x$ or $xy = k$. The graph is a hyperbola.
* Graph A: This is a straight line.
* Graph B: This is a curve that looks like a hyperbola branch. It passes through the point $(2, 4)$.
* Conclusion: Graph B represents inverse variation.
* Finding the equation: Using the point $(2, 4)$:
$$y = k/x$$
$$4 = k/2$$
$$k = 8$$
So, the equation is $y = 8/x$ (or $xy = 8$).

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Section 2: Identify Direct or Inverse Variation

Problem 3: "The total cost of apples is $2 per pound."
* Analysis: Total Cost ($C$) = Price per pound ($2$) $\times$ Weight ($w$).
* Equation: $C = 2w$.
* Conclusion: This is a Direct Variation ($y = kx$).

Problem 4: "The time it takes to drive a fixed distance is inversely proportional to speed."
* Analysis: The problem explicitly states "inversely proportional".
* Equation: $Time = Distance / Speed$.
* Conclusion: This is an Inverse Variation.

Problem 5: "The area of a square is directly proportional to the square of its side length."
* Analysis: Area ($A$) = side ($s$) $\times$ side ($s$) = $s^2$.
* Equation: $A = 1 \cdot s^2$.
* Conclusion: While strictly speaking this is a variation involving a power, in the context of "Direct vs Inverse", it falls under Direct Variation (as $A$ increases when $s$ increases). Specifically, $A$ varies directly as $s^2$.

Problem 6: "The number of people sharing a pizza is inversely proportional to the amount of pizza each person gets."
* Analysis: If $P$ is people and $A$ is amount per person, and Pizza Size is constant ($S$), then $P \times A = S$.
* Equation: $A = S / P$.
* Conclusion: This is an Inverse Variation.

Problem 7: "The distance traveled at a constant speed is directly proportional to the time traveled."
* Analysis: Distance ($d$) = Speed ($r$) $\times$ Time ($t$).
* Equation: $d = rt$.
* Conclusion: This is a Direct Variation.

Problem 8: "The intensity of light is inversely proportional to the square of the distance from the source."
* Analysis: Intensity ($I$) varies as $1/d^2$.
* Equation: $I = k / d^2$.
* Conclusion: This is an Inverse Variation.

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Section 3: Write an Inverse Variation Equation

Problem 9: "$y$ varies inversely as $x$. If $x = 4$ when $y = 10$, find $y$ when $x = 8$."
* Step 1: Find $k$.
$$y = k/x$$
$$10 = k/4$$
$$k = 40$$
Equation: $y = 40/x$
* Step 2: Find $y$ when $x = 8$.
$$y = 40/8$$
$$y = 5$$

Problem 10: "$y$ varies inversely as $x$. If $x = 6$ when $y = 9$, find $y$ when $x = 18$."
* Step 1: Find $k$.
$$y = k/x$$
$$9 = k/6$$
$$k = 54$$
Equation: $y = 54/x$
* Step 2: Find $y$ when $x = 18$.
$$y = 54/18$$
$$y = 3$$

Problem 11: "$y$ varies inversely as $x$. If $x = 12$ when $y = 5$, find $y$ when $x = 20$."
* Step 1: Find $k$.
$$y = k/x$$
$$5 = k/12$$
$$k = 60$$
Equation: $y = 60/x$
* Step 2: Find $y$ when $x = 20$.
$$y = 60/20$$
$$y = 3$$

Problem 12: "$y$ varies inversely as $x$. If $x = 3$ when $y = 15$, find $y$ when $x = 9$."
* Step 1: Find $k$.
$$y = k/x$$
$$15 = k/3$$
$$k = 45$$
Equation: $y = 45/x$
* Step 2: Find $y$ when $x = 9$.
$$y = 45/9$$
$$y = 5$$

Problem 13: "$y$ varies inversely as $x$. If $x = 5$ when $y = 20$, find $y$ when $x = 25$."
* Step 1: Find $k$.
$$y = k/x$$
$$20 = k/5$$
$$k = 100$$
Equation: $y = 100/x$
* Step 2: Find $y$ when $x = 25$.
$$y = 100/25$$
$$y = 4$$

Problem 14: "$y$ varies inversely as $x$. If $x = 8$ when $y = 12$, find $y$ when $x = 6$."
* Step 1: Find $k$.
$$y = k/x$$
$$12 = k/8$$
$$k = 96$$
Equation: $y = 96/x$
* Step 2: Find $y$ when $x = 6$.
$$y = 96/6$$
$$y = 16$$

Problem 15: "$y$ varies inversely as $x$. If $x = 10$ when $y = 4$, find $y$ when $x = 20$."
* Step 1: Find $k$.
$$y = k/x$$
$$4 = k/10$$
$$k = 40$$
Equation: $y = 40/x$
* Step 2: Find $y$ when $x = 20$.
$$y = 40/20$$
$$y = 2$$

Problem 16: "$y$ varies inversely as $x$. If $x = 7$ when $y = 14$, find $y$ when $x = 2$."
* Step 1: Find $k$.
$$y = k/x$$
$$14 = k/7$$
$$k = 98$$
Equation: $y = 98/x$
* Step 2: Find $y$ when $x = 2$.
$$y = 98/2$$
$$y = 49$$

Problem 17: "$y$ varies inversely as $x$. If $x = 4$ when $y = 16$, find $y$ when $x = 8$."
* Step 1: Find $k$.
$$y = k/x$$
$$16 = k/4$$
$$k = 64$$
Equation: $y = 64/x$
* Step 2: Find $y$ when $x = 8$.
$$y = 64/8$$
$$y = 8$$

Problem 18: "$y$ varies inversely as $x$. If $x = 9$ when $y = 3$, find $y$ when $x = 27$."
* Step 1: Find $k$.
$$y = k/x$$
$$3 = k/9$$
$$k = 27$$
Equation: $y = 27/x$
* Step 2: Find $y$ when $x = 27$.
$$y = 27/27$$
$$y = 1$$

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Final Answer:

Graphs Section:
1. Graph A represents direct variation. Equation: $y = 2x$
2. Graph B represents inverse variation. Equation: $y = 8/x$

Identify Variation Section:
3. Direct Variation
4. Inverse Variation
5. Direct Variation
6. Inverse Variation
7. Direct Variation
8. Inverse Variation

Write Equation & Solve Section:
9. $y = 5$
10. $y = 3$
11. $y = 3$
12. $y = 5$
13. $y = 4$
14. $y = 16$
15. $y = 2$
16. $y = 49$
17. $y = 8$
18. $y = 1$