Examples of solving rational equations with variables in denominators.

To solve the given equations, we will address each one step by step. Let's go through them systematically.

---

Equation 1:

\[
\frac{7}{x} = \frac{3}{2x + 5}
\]

#### Step 1: Cross-multiply to eliminate the fractions.
\[
7(2x + 5) = 3x
\]

#### Step 2: Expand and simplify.
\[
14x + 35 = 3x
\]

#### Step 3: Isolate \( x \).
Subtract \( 3x \) from both sides:
\[
14x - 3x + 35 = 0
\]
\[
11x + 35 = 0
\]

Subtract 35 from both sides:
\[
11x = -35
\]

Divide by 11:
\[
x = -\frac{35}{11}
\]

#### Step 4: Verify the solution.
Substitute \( x = -\frac{35}{11} \) back into the original equation to ensure it satisfies both sides:
\[
\frac{7}{-\frac{35}{11}} = \frac{3}{2\left(-\frac{35}{11}\right) + 5}
\]

Simplify the left-hand side:
\[
\frac{7}{-\frac{35}{11}} = 7 \cdot \left(-\frac{11}{35}\right) = -\frac{77}{35} = -\frac{11}{5}
\]

Simplify the right-hand side:
\[
2\left(-\frac{35}{11}\right) + 5 = -\frac{70}{11} + \frac{55}{11} = -\frac{15}{11}
\]
\[
\frac{3}{-\frac{15}{11}} = 3 \cdot \left(-\frac{11}{15}\right) = -\frac{33}{15} = -\frac{11}{5}
\]

Both sides are equal, so the solution is verified.

\[
\boxed{x = -\frac{35}{11}}
\]

---

Equation 2:

\[
\frac{7}{x} - \frac{4}{2x} = \frac{3}{2x + 5}
\]

#### Step 1: Simplify the left-hand side.
\[
\frac{7}{x} - \frac{4}{2x} = \frac{7}{x} - \frac{2}{x} = \frac{5}{x}
\]

So the equation becomes:
\[
\frac{5}{x} = \frac{3}{2x + 5}
\]

#### Step 2: Cross-multiply to eliminate the fractions.
\[
5(2x + 5) = 3x
\]

#### Step 3: Expand and simplify.
\[
10x + 25 = 3x
\]

#### Step 4: Isolate \( x \).
Subtract \( 3x \) from both sides:
\[
10x - 3x + 25 = 0
\]
\[
7x + 25 = 0
\]

Subtract 25 from both sides:
\[
7x = -25
\]

Divide by 7:
\[
x = -\frac{25}{7}
\]

#### Step 5: Verify the solution.
Substitute \( x = -\frac{25}{7} \) back into the original equation to ensure it satisfies both sides:
\[
\frac{7}{-\frac{25}{7}} - \frac{4}{2\left(-\frac{25}{7}\right)} = \frac{3}{2\left(-\frac{25}{7}\right) + 5}
\]

Simplify the left-hand side:
\[
\frac{7}{-\frac{25}{7}} = 7 \cdot \left(-\frac{7}{25}\right) = -\frac{49}{25}
\]
\[
\frac{4}{2\left(-\frac{25}{7}\right)} = \frac{4}{-\frac{50}{7}} = 4 \cdot \left(-\frac{7}{50}\right) = -\frac{28}{50} = -\frac{14}{25}
\]
\[
\frac{7}{-\frac{25}{7}} - \frac{4}{2\left(-\frac{25}{7}\right)} = -\frac{49}{25} - \left(-\frac{14}{25}\right) = -\frac{49}{25} + \frac{14}{25} = -\frac{35}{25} = -\frac{7}{5}
\]

Simplify the right-hand side:
\[
2\left(-\frac{25}{7}\right) + 5 = -\frac{50}{7} + \frac{35}{7} = -\frac{15}{7}
\]
\[
\frac{3}{-\frac{15}{7}} = 3 \cdot \left(-\frac{7}{15}\right) = -\frac{21}{15} = -\frac{7}{5}
\]

Both sides are equal, so the solution is verified.

\[
\boxed{x = -\frac{25}{7}}
\]

---

Equation 3:

\[
\frac{7}{1 - 6x} = \frac{3}{2x + 5}
\]

#### Step 1: Cross-multiply to eliminate the fractions.
\[
7(2x + 5) = 3(1 - 6x)
\]

#### Step 2: Expand and simplify.
\[
14x + 35 = 3 - 18x
\]

#### Step 3: Isolate \( x \).
Add \( 18x \) to both sides:
\[
14x + 18x + 35 = 3
\]
\[
32x + 35 = 3
\]

Subtract 35 from both sides:
\[
32x = 3 - 35
\]
\[
32x = -32
\]

Divide by 32:
\[
x = -1
\]

#### Step 4: Verify the solution.
Substitute \( x = -1 \) back into the original equation to ensure it satisfies both sides:
\[
\frac{7}{1 - 6(-1)} = \frac{3}{2(-1) + 5}
\]

Simplify the left-hand side:
\[
1 - 6(-1) = 1 + 6 = 7
\]
\[
\frac{7}{7} = 1
\]

Simplify the right-hand side:
\[
2(-1) + 5 = -2 + 5 = 3
\]
\[
\frac{3}{3} = 1
\]

Both sides are equal, so the solution is verified.

\[
\boxed{x = -1}
\]

---

Equation 4:

\[
\frac{7}{1 - 6x} = \frac{3}{2(x + 5)}
\]

#### Step 1: Cross-multiply to eliminate the fractions.
\[
7 \cdot 2(x + 5) = 3(1 - 6x)
\]

#### Step 2: Expand and simplify.
\[
14(x + 5) = 3 - 18x
\]
\[
14x + 70 = 3 - 18x
\]

#### Step 3: Isolate \( x \).
Add \( 18x \) to both sides:
\[
14x + 18x + 70 = 3
\]
\[
32x + 70 = 3
\]

Subtract 70 from both sides:
\[
32x = 3 - 70
\]
\[
32x = -67
\]

Divide by 32:
\[
x = -\frac{67}{32}
\]

#### Step 4: Verify the solution.
Substitute \( x = -\frac{67}{32} \) back into the original equation to ensure it satisfies both sides:
\[
\frac{7}{1 - 6\left(-\frac{67}{32}\right)} = \frac{3}{2\left(-\frac{67}{32} + 5\right)}
\]

Simplify the left-hand side:
\[
1 - 6\left(-\frac{67}{32}\right) = 1 + \frac{402}{32} = \frac{32}{32} + \frac{402}{32} = \frac{434}{32} = \frac{217}{16}
\]
\[
\frac{7}{\frac{217}{16}} = 7 \cdot \frac{16}{217} = \frac{112}{217}
\]

Simplify the right-hand side:
\[
-\frac{67}{32} + 5 = -\frac{67}{32} + \frac{160}{32} = \frac{93}{32}
\]
\[
2\left(\frac{93}{32}\right) = \frac{186}{32} = \frac{93}{16}
\]
\[
\frac{3}{\frac{93}{16}} = 3 \cdot \frac{16}{93} = \frac{48}{93} = \frac{16}{31}
\]

Both sides are equal, so the solution is verified.

\[
\boxed{x = -\frac{67}{32}}
\]

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Final Answers:

\[
\boxed{-\frac{35}{11}, -\frac{25}{7}, -1, -\frac{67}{32}}
\]