Solving Rational Equations Worksheet featuring ten algebraic problems and a cartoon teacher illustration.

Solving Rational Equations Worksheet

We will solve each equation step by step and check for excluded values. Let's go through the problems one by one.

---

#### Problem 1:
$$
\frac{2a - 3}{6} = \frac{2a}{3} + \frac{1}{2}
$$

Step 1: Find a common denominator.
The denominators are 6, 3, and 2. The least common denominator (LCD) is 6.

Step 2: Rewrite each term with the LCD.
$$
\frac{2a - 3}{6} = \frac{2a}{3} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{3}{3}
$$
$$
\frac{2a - 3}{6} = \frac{4a}{6} + \frac{3}{6}
$$

Step 3: Combine the terms on the right-hand side.
$$
\frac{2a - 3}{6} = \frac{4a + 3}{6}
$$

Step 4: Eliminate the denominators by multiplying both sides by 6.
$$
2a - 3 = 4a + 3
$$

Step 5: Solve for \(a\).
$$
2a - 4a = 3 + 3
$$
$$
-2a = 6
$$
$$
a = -3
$$

Step 6: Check for excluded values.
The original equation has denominators of 6, 3, and 2, which are all constants and do not depend on \(a\). Therefore, there are no excluded values.

Final Answer:
$$
\boxed{-3}
$$

---

#### Problem 2:
$$
\frac{2b - 3}{7} - \frac{b}{2} = \frac{b + 3}{14}
$$

Step 1: Find a common denominator.
The denominators are 7, 2, and 14. The LCD is 14.

Step 2: Rewrite each term with the LCD.
$$
\frac{2b - 3}{7} \cdot \frac{2}{2} - \frac{b}{2} \cdot \frac{7}{7} = \frac{b + 3}{14}
$$
$$
\frac{2(2b - 3)}{14} - \frac{7b}{14} = \frac{b + 3}{14}
$$
$$
\frac{4b - 6}{14} - \frac{7b}{14} = \frac{b + 3}{14}
$$

Step 3: Combine the terms on the left-hand side.
$$
\frac{4b - 6 - 7b}{14} = \frac{b + 3}{14}
$$
$$
\frac{-3b - 6}{14} = \frac{b + 3}{14}
$$

Step 4: Eliminate the denominators by multiplying both sides by 14.
$$
-3b - 6 = b + 3
$$

Step 5: Solve for \(b\).
$$
-3b - b = 3 + 6
$$
$$
-4b = 9
$$
$$
b = -\frac{9}{4}
$$

Step 6: Check for excluded values.
The original equation has denominators of 7, 2, and 14, which are all constants and do not depend on \(b\). Therefore, there are no excluded values.

Final Answer:
$$
\boxed{-\frac{9}{4}}
$$

---

#### Problem 3:
$$
\frac{3}{5x} + \frac{7}{2x} = 1
$$

Step 1: Find a common denominator.
The denominators are \(5x\) and \(2x\). The LCD is \(10x\).

Step 2: Rewrite each term with the LCD.
$$
\frac{3}{5x} \cdot \frac{2}{2} + \frac{7}{2x} \cdot \frac{5}{5} = 1
$$
$$
\frac{6}{10x} + \frac{35}{10x} = 1
$$

Step 3: Combine the terms on the left-hand side.
$$
\frac{6 + 35}{10x} = 1
$$
$$
\frac{41}{10x} = 1
$$

Step 4: Eliminate the denominator by multiplying both sides by \(10x\).
$$
41 = 10x
$$

Step 5: Solve for \(x\).
$$
x = \frac{41}{10}
$$

Step 6: Check for excluded values.
The original equation has denominators of \(5x\) and \(2x\). These are zero when \(x = 0\). Therefore, \(x = 0\) is an excluded value.

Final Answer:
$$
\boxed{\frac{41}{10}}
$$

---

#### Problem 4:
$$
\frac{5k}{k + 2} + \frac{2}{k} = 5
$$

Step 1: Find a common denominator.
The denominators are \(k + 2\) and \(k\). The LCD is \(k(k + 2)\).

Step 2: Rewrite each term with the LCD.
$$
\frac{5k}{k + 2} \cdot \frac{k}{k} + \frac{2}{k} \cdot \frac{k + 2}{k + 2} = 5
$$
$$
\frac{5k^2}{k(k + 2)} + \frac{2(k + 2)}{k(k + 2)} = 5
$$

Step 3: Combine the terms on the left-hand side.
$$
\frac{5k^2 + 2(k + 2)}{k(k + 2)} = 5
$$
$$
\frac{5k^2 + 2k + 4}{k(k + 2)} = 5
$$

Step 4: Eliminate the denominator by multiplying both sides by \(k(k + 2)\).
$$
5k^2 + 2k + 4 = 5k(k + 2)
$$

Step 5: Expand the right-hand side.
$$
5k^2 + 2k + 4 = 5k^2 + 10k
$$

Step 6: Simplify the equation.
$$
5k^2 + 2k + 4 - 5k^2 - 10k = 0
$$
$$
-8k + 4 = 0
$$

Step 7: Solve for \(k\).
$$
-8k = -4
$$
$$
k = \frac{1}{2}
$$

Step 8: Check for excluded values.
The original equation has denominators of \(k + 2\) and \(k\). These are zero when \(k = -2\) or \(k = 0\). Therefore, \(k = -2\) and \(k = 0\) are excluded values.

Final Answer:
$$
\boxed{\frac{1}{2}}
$$

---

#### Problem 5:
$$
\frac{m}{m + 1} + \frac{5}{m - 1} = 1
$$

Step 1: Find a common denominator.
The denominators are \(m + 1\) and \(m - 1\). The LCD is \((m + 1)(m - 1)\).

Step 2: Rewrite each term with the LCD.
$$
\frac{m}{m + 1} \cdot \frac{m - 1}{m - 1} + \frac{5}{m - 1} \cdot \frac{m + 1}{m + 1} = 1
$$
$$
\frac{m(m - 1)}{(m + 1)(m - 1)} + \frac{5(m + 1)}{(m + 1)(m - 1)} = 1
$$

Step 3: Combine the terms on the left-hand side.
$$
\frac{m(m - 1) + 5(m + 1)}{(m + 1)(m - 1)} = 1
$$
$$
\frac{m^2 - m + 5m + 5}{(m + 1)(m - 1)} = 1
$$
$$
\frac{m^2 + 4m + 5}{(m + 1)(m - 1)} = 1
$$

Step 4: Eliminate the denominator by multiplying both sides by \((m + 1)(m - 1)\).
$$
m^2 + 4m + 5 = (m + 1)(m - 1)
$$

Step 5: Expand the right-hand side.
$$
m^2 + 4m + 5 = m^2 - 1
$$

Step 6: Simplify the equation.
$$
m^2 + 4m + 5 - m^2 + 1 = 0
$$
$$
4m + 6 = 0
$$

Step 7: Solve for \(m\).
$$
4m = -6
$$
$$
m = -\frac{3}{2}
$$

Step 8: Check for excluded values.
The original equation has denominators of \(m + 1\) and \(m - 1\). These are zero when \(m = -1\) or \(m = 1\). Therefore, \(m = -1\) and \(m = 1\) are excluded values.

Final Answer:
$$
\boxed{-\frac{3}{2}}
$$

---

Final Answers:

1. \(\boxed{-3}\)
2. \(\boxed{-\frac{9}{4}}\)
3. \(\boxed{\frac{41}{10}}\)
4. \(\boxed{\frac{1}{2}}\)
5. \(\boxed{-\frac{3}{2}}\)

I will continue solving the remaining problems if needed. Let me know!