Solve Proportions Worksheets [PDF] (7.RP.A.2.C): 7th Grade Math - Free Printable

Problem Overview:

The worksheet involves solving problems related to inverse proportion and inverse proportion to powers of \( x \). Inverse proportion means that as one variable increases, the other decreases in such a way that their product is constant. Mathematically, if \( y \) is inversely proportional to \( x \), then:

\[
y = \frac{k}{x}
\]

where \( k \) is the constant of proportionality.

For inverse proportion to powers of \( x \):
- If \( y \) is inversely proportional to \( x^2 \), then \( y = \frac{k}{x^2} \).
- If \( y \) is inversely proportional to \( x^3 \), then \( y = \frac{k}{x^3} \).

We will solve each table step by step and determine the missing values using these relationships.

---

Section 1: \( y \) is inversely proportional to \( x \)

#### Table 1:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
y & ? & 6 & ? & ? & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 6 \) when \( x = 2 \):
\[
k = x \cdot y = 2 \cdot 6 = 12
\]

2. Fill in the table:
- For \( x = 1 \):
\[
y = \frac{k}{x} = \frac{12}{1} = 12
\]
- For \( x = 3 \):
\[
y = \frac{k}{x} = \frac{12}{3} = 4
\]
- For \( x = 4 \):
\[
y = \frac{k}{x} = \frac{12}{4} = 3
\]
- For \( x = 5 \):
\[
y = \frac{k}{x} = \frac{12}{5} = 2.4
\]
- For \( x = 6 \):
\[
y = \frac{k}{x} = \frac{12}{6} = 2
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
y & 12 & 6 & 4 & 3 & 2.4 & 2
\end{array}
\]

#### Table 2:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
y & ? & ? & 18 & ? & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 18 \) when \( x = 3 \):
\[
k = x \cdot y = 3 \cdot 18 = 54
\]

2. Fill in the table:
- For \( x = 1 \):
\[
y = \frac{k}{x} = \frac{54}{1} = 54
\]
- For \( x = 2 \):
\[
y = \frac{k}{x} = \frac{54}{2} = 27
\]
- For \( x = 4 \):
\[
y = \frac{k}{x} = \frac{54}{4} = 13.5
\]
- For \( x = 5 \):
\[
y = \frac{k}{x} = \frac{54}{5} = 10.8
\]
- For \( x = 6 \):
\[
y = \frac{k}{x} = \frac{54}{6} = 9
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
y & 54 & 27 & 18 & 13.5 & 10.8 & 9
\end{array}
\]

#### Table 3:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 4 & 8 & 10 & 12 \\
\hline
y & ? & ? & ? & 35 & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 35 \) when \( x = 8 \):
\[
k = x \cdot y = 8 \cdot 35 = 280
\]

2. Fill in the table:
- For \( x = 1 \):
\[
y = \frac{k}{x} = \frac{280}{1} = 280
\]
- For \( x = 2 \):
\[
y = \frac{k}{x} = \frac{280}{2} = 140
\]
- For \( x = 4 \):
\[
y = \frac{k}{x} = \frac{280}{4} = 70
\]
- For \( x = 10 \):
\[
y = \frac{k}{x} = \frac{280}{10} = 28
\]
- For \( x = 12 \):
\[
y = \frac{k}{x} = \frac{280}{12} \approx 23.33
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & 1 & 2 & 4 & 8 & 10 & 12 \\
\hline
y & 280 & 140 & 70 & 35 & 28 & 23.33
\end{array}
\]

---

Section 2: \( y \) is inversely proportional to \( x \) (with negative values)

#### Table 4:
\[
\begin{array}{c|c|c|c|c|c}
x & -3 & -2 & -1 & 1 & 2 & 3 \\
\hline
y & ? & ? & ? & 8 & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 8 \) when \( x = 1 \):
\[
k = x \cdot y = 1 \cdot 8 = 8
\]

2. Fill in the table:
- For \( x = -3 \):
\[
y = \frac{k}{x} = \frac{8}{-3} \approx -2.67
\]
- For \( x = -2 \):
\[
y = \frac{k}{x} = \frac{8}{-2} = -4
\]
- For \( x = -1 \):
\[
y = \frac{k}{x} = \frac{8}{-1} = -8
\]
- For \( x = 2 \):
\[
y = \frac{k}{x} = \frac{8}{2} = 4
\]
- For \( x = 3 \):
\[
y = \frac{k}{x} = \frac{8}{3} \approx 2.67
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -3 & -2 & -1 & 1 & 2 & 3 \\
\hline
y & -2.67 & -4 & -8 & 8 & 4 & 2.67
\end{array}
\]

#### Table 5:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & ? & ? & ? & 15 & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 15 \) when \( x = 2 \):
\[
k = x \cdot y = 2 \cdot 15 = 30
\]

2. Fill in the table:
- For \( x = -6 \):
\[
y = \frac{k}{x} = \frac{30}{-6} = -5
\]
- For \( x = -4 \):
\[
y = \frac{k}{x} = \frac{30}{-4} = -7.5
\]
- For \( x = -2 \):
\[
y = \frac{k}{x} = \frac{30}{-2} = -15
\]
- For \( x = 4 \):
\[
y = \frac{k}{x} = \frac{30}{4} = 7.5
\]
- For \( x = 6 \):
\[
y = \frac{k}{x} = \frac{30}{6} = 5
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & -5 & -7.5 & -15 & 15 & 7.5 & 5
\end{array}
\]

#### Table 6:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & ? & ? & ? & 10 & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 10 \) when \( x = 2 \):
\[
k = x \cdot y = 2 \cdot 10 = 20
\]

2. Fill in the table:
- For \( x = -6 \):
\[
y = \frac{k}{x} = \frac{20}{-6} \approx -3.33
\]
- For \( x = -4 \):
\[
y = \frac{k}{x} = \frac{20}{-4} = -5
\]
- For \( x = -2 \):
\[
y = \frac{k}{x} = \frac{20}{-2} = -10
\]
- For \( x = 4 \):
\[
y = \frac{k}{x} = \frac{20}{4} = 5
\]
- For \( x = 6 \):
\[
y = \frac{k}{x} = \frac{20}{6} \approx 3.33
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & -3.33 & -5 & -10 & 10 & 5 & 3.33
\end{array}
\]

---

Section 3: \( y \) is inversely proportional to \( x^2 \) and \( x^3 \)

#### Table 7:
\[
\begin{array}{c|c|c|c|c|c}
x & -3 & -2 & -1 & 1 & 2 & 3 \\
\hline
y & ? & ? & ? & ? & 9 & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 9 \) when \( x = 2 \):
\[
k = x^2 \cdot y = 2^2 \cdot 9 = 4 \cdot 9 = 36
\]

2. Fill in the table:
- For \( x = -3 \):
\[
y = \frac{k}{x^2} = \frac{36}{(-3)^2} = \frac{36}{9} = 4
\]
- For \( x = -2 \):
\[
y = \frac{k}{x^2} = \frac{36}{(-2)^2} = \frac{36}{4} = 9
\]
- For \( x = -1 \):
\[
y = \frac{k}{x^2} = \frac{36}{(-1)^2} = \frac{36}{1} = 36
\]
- For \( x = 1 \):
\[
y = \frac{k}{x^2} = \frac{36}{1^2} = \frac{36}{1} = 36
\]
- For \( x = 3 \):
\[
y = \frac{k}{x^2} = \frac{36}{3^2} = \frac{36}{9} = 4
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -3 & -2 & -1 & 1 & 2 & 3 \\
\hline
y & 4 & 9 & 36 & 36 & 9 & 4
\end{array}
\]

#### Table 8:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & ? & ? & \frac{16}{9} & ? & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = \frac{16}{9} \) when \( x = -2 \):
\[
k = x^2 \cdot y = (-2)^2 \cdot \frac{16}{9} = 4 \cdot \frac{16}{9} = \frac{64}{9}
\]

2. Fill in the table:
- For \( x = -6 \):
\[
y = \frac{k}{x^2} = \frac{\frac{64}{9}}{(-6)^2} = \frac{\frac{64}{9}}{36} = \frac{64}{9 \cdot 36} = \frac{64}{324} = \frac{16}{81}
\]
- For \( x = -4 \):
\[
y = \frac{k}{x^2} = \frac{\frac{64}{9}}{(-4)^2} = \frac{\frac{64}{9}}{16} = \frac{64}{9 \cdot 16} = \frac{64}{144} = \frac{4}{9}
\]
- For \( x = 2 \):
\[
y = \frac{k}{x^2} = \frac{\frac{64}{9}}{2^2} = \frac{\frac{64}{9}}{4} = \frac{64}{9 \cdot 4} = \frac{64}{36} = \frac{16}{9}
\]
- For \( x = 4 \):
\[
y = \frac{k}{x^2} = \frac{\frac{64}{9}}{4^2} = \frac{\frac{64}{9}}{16} = \frac{64}{9 \cdot 16} = \frac{64}{144} = \frac{4}{9}
\]
- For \( x = 6 \):
\[
y = \frac{k}{x^2} = \frac{\frac{64}{9}}{6^2} = \frac{\frac{64}{9}}{36} = \frac{64}{9 \cdot 36} = \frac{64}{324} = \frac{16}{81}
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & \frac{16}{81} & \frac{4}{9} & \frac{16}{9} & \frac{16}{9} & \frac{4}{9} & \frac{16}{81}
\end{array}
\]

#### Table 9:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & ? & ? & ? & 25 & ? & ?
\end{array}
\]

1. Find \( k \):
- Given \( y = 25 \) when \( x = 2 \):
\[
k = x^3 \cdot y = 2^3 \cdot 25 = 8 \cdot 25 = 200
\]

2. Fill in the table:
- For \( x = -6 \):
\[
y = \frac{k}{x^3} = \frac{200}{(-6)^3} = \frac{200}{-216} = -\frac{25}{27}
\]
- For \( x = -4 \):
\[
y = \frac{k}{x^3} = \frac{200}{(-4)^3} = \frac{200}{-64} = -\frac{25}{8}
\]
- For \( x = -2 \):
\[
y = \frac{k}{x^3} = \frac{200}{(-2)^3} = \frac{200}{-8} = -25
\]
- For \( x = 4 \):
\[
y = \frac{k}{x^3} = \frac{200}{4^3} = \frac{200}{64} = \frac{25}{8}
\]
- For \( x = 6 \):
\[
y = \frac{k}{x^3} = \frac{200}{6^3} = \frac{200}{216} = \frac{25}{27}
\]

Completed Table:
\[
\begin{array}{c|c|c|c|c|c}
x & -6 & -4 & -2 & 2 & 4 & 6 \\
\hline
y & -\frac{25}{27} & -\frac{25}{8} & -25 & 25 & \frac{25}{8} & \frac{25}{27}
\end{array}
\]

---

Final Answer:

\[
\boxed{
\begin{array}{c}
\text{Tables filled as shown above.}
\end{array}
}
\]