Worksheet #10 features 15 linear equation problems for Mathematics Pure Module A students to solve without calculators.

Problem: Solve each linear equation in one variable.

The worksheet provides 15 linear equations, and we are tasked with solving them step by step. Below is the detailed solution for each equation.

---

#### Equation 1: \( a + 7a + 5 = -5a + 5 \)

1. Combine like terms on the left-hand side:
\[
a + 7a + 5 = 8a + 5
\]
2. The equation becomes:
\[
8a + 5 = -5a + 5
\]
3. Subtract 5 from both sides:
\[
8a = -5a
\]
4. Add \( 5a \) to both sides:
\[
8a + 5a = 0
\]
\[
13a = 0
\]
5. Divide by 13:
\[
a = 0
\]

Solution: \( \boxed{a = 0} \)

---

#### Equation 2: \( p - 1 = 5p + 3p - 8 \)

1. Combine like terms on the right-hand side:
\[
5p + 3p = 8p
\]
So the equation becomes:
\[
p - 1 = 8p - 8
\]
2. Subtract \( p \) from both sides:
\[
-1 = 7p - 8
\]
3. Add 8 to both sides:
\[
7 = 7p
\]
4. Divide by 7:
\[
p = 1
\]

Solution: \( \boxed{p = 1} \)

---

#### Equation 3: \( 2(4x - 3) - 8 = 4 + 2x \)

1. Distribute the 2 on the left-hand side:
\[
2(4x - 3) = 8x - 6
\]
So the equation becomes:
\[
8x - 6 - 8 = 4 + 2x
\]
2. Simplify the left-hand side:
\[
8x - 14 = 4 + 2x
\]
3. Subtract \( 2x \) from both sides:
\[
8x - 2x - 14 = 4
\]
\[
6x - 14 = 4
\]
4. Add 14 to both sides:
\[
6x = 18
\]
5. Divide by 6:
\[
x = 3
\]

Solution: \( \boxed{x = 3} \)

---

#### Equation 4: \( 3n - 5 = -8(6 + 5n) \)

1. Distribute the \(-8\) on the right-hand side:
\[
-8(6 + 5n) = -48 - 40n
\]
So the equation becomes:
\[
3n - 5 = -48 - 40n
\]
2. Add \( 40n \) to both sides:
\[
3n + 40n - 5 = -48
\]
\[
43n - 5 = -48
\]
3. Add 5 to both sides:
\[
43n = -43
\]
4. Divide by 43:
\[
n = -1
\]

Solution: \( \boxed{n = -1} \)

---

#### Equation 5: \( 24a - 22 = -4(1 - 6a) \)

1. Distribute the \(-4\) on the right-hand side:
\[
-4(1 - 6a) = -4 + 24a
\]
So the equation becomes:
\[
24a - 22 = -4 + 24a
\]
2. Subtract \( 24a \) from both sides:
\[
-22 = -4
\]
This is a contradiction, so there is no solution.

Solution: \( \boxed{\text{No solution}} \)

---

#### Equation 6: \( -3(4x + 3) + 4(6x + 1) = 43 \)

1. Distribute the constants:
\[
-3(4x + 3) = -12x - 9
\]
\[
4(6x + 1) = 24x + 4
\]
So the equation becomes:
\[
-12x - 9 + 24x + 4 = 43
\]
2. Combine like terms:
\[
(-12x + 24x) + (-9 + 4) = 43
\]
\[
12x - 5 = 43
\]
3. Add 5 to both sides:
\[
12x = 48
\]
4. Divide by 12:
\[
x = 4
\]

Solution: \( \boxed{x = 4} \)

---

#### Equation 7: \( \frac{3a + 2}{5} = \frac{7 - 2a}{2} \)

1. Eliminate the fractions by cross-multiplying:
\[
2(3a + 2) = 5(7 - 2a)
\]
2. Distribute on both sides:
\[
2(3a + 2) = 6a + 4
\]
\[
5(7 - 2a) = 35 - 10a
\]
So the equation becomes:
\[
6a + 4 = 35 - 10a
\]
3. Add \( 10a \) to both sides:
\[
6a + 10a + 4 = 35
\]
\[
16a + 4 = 35
\]
4. Subtract 4 from both sides:
\[
16a = 31
\]
5. Divide by 16:
\[
a = \frac{31}{16}
\]

Solution: \( \boxed{a = \frac{31}{16}} \)

---

#### Equation 8: \( \frac{x}{2} + \frac{2x - 1}{3} = \frac{3x + 4}{4} \)

1. Find a common denominator for all fractions, which is 12:
\[
\frac{x}{2} = \frac{6x}{12}, \quad \frac{2x - 1}{3} = \frac{4(2x - 1)}{12} = \frac{8x - 4}{12}, \quad \frac{3x + 4}{4} = \frac{3(3x + 4)}{12} = \frac{9x + 12}{12}
\]
So the equation becomes:
\[
\frac{6x}{12} + \frac{8x - 4}{12} = \frac{9x + 12}{12}
\]
2. Combine the fractions on the left-hand side:
\[
\frac{6x + (8x - 4)}{12} = \frac{9x + 12}{12}
\]
\[
\frac{14x - 4}{12} = \frac{9x + 12}{12}
\]
3. Since the denominators are the same, equate the numerators:
\[
14x - 4 = 9x + 12
\]
4. Subtract \( 9x \) from both sides:
\[
14x - 9x - 4 = 12
\]
\[
5x - 4 = 12
\]
5. Add 4 to both sides:
\[
5x = 16
\]
6. Divide by 5:
\[
x = \frac{16}{5}
\]

Solution: \( \boxed{x = \frac{16}{5}} \)

---

#### Equation 9: \( \frac{x}{5} + \frac{3x - 1}{2} = \frac{6x + 5}{4} \)

1. Find a common denominator for all fractions, which is 20:
\[
\frac{x}{5} = \frac{4x}{20}, \quad \frac{3x - 1}{2} = \frac{10(3x - 1)}{20} = \frac{30x - 10}{20}, \quad \frac{6x + 5}{4} = \frac{5(6x + 5)}{20} = \frac{30x + 25}{20}
\]
So the equation becomes:
\[
\frac{4x}{20} + \frac{30x - 10}{20} = \frac{30x + 25}{20}
\]
2. Combine the fractions on the left-hand side:
\[
\frac{4x + (30x - 10)}{20} = \frac{30x + 25}{20}
\]
\[
\frac{34x - 10}{20} = \frac{30x + 25}{20}
\]
3. Since the denominators are the same, equate the numerators:
\[
34x - 10 = 30x + 25
\]
4. Subtract \( 30x \) from both sides:
\[
34x - 30x - 10 = 25
\]
\[
4x - 10 = 25
\]
5. Add 10 to both sides:
\[
4x = 35
\]
6. Divide by 4:
\[
x = \frac{35}{4}
\]

Solution: \( \boxed{x = \frac{35}{4}} \)

---

#### Equation 10: \( \frac{2m}{5} + \frac{m - 4}{6} = \frac{4m + 1}{4} - 2 \)

1. Find a common denominator for all fractions, which is 60:
\[
\frac{2m}{5} = \frac{24m}{60}, \quad \frac{m - 4}{6} = \frac{10(m - 4)}{60} = \frac{10m - 40}{60}, \quad \frac{4m + 1}{4} = \frac{15(4m + 1)}{60} = \frac{60m + 15}{60}
\]
So the equation becomes:
\[
\frac{24m}{60} + \frac{10m - 40}{60} = \frac{60m + 15}{60} - 2
\]
2. Combine the fractions on the left-hand side:
\[
\frac{24m + (10m - 40)}{60} = \frac{60m + 15}{60} - 2
\]
\[
\frac{34m - 40}{60} = \frac{60m + 15}{60} - 2
\]
3. Eliminate the fraction by multiplying through by 60:
\[
34m - 40 = 60m + 15 - 120
\]
\[
34m - 40 = 60m - 105
\]
4. Subtract \( 34m \) from both sides:
\[
-40 = 26m - 105
\]
5. Add 105 to both sides:
\[
65 = 26m
\]
6. Divide by 26:
\[
m = \frac{65}{26} = \frac{5}{2}
\]

Solution: \( \boxed{m = \frac{5}{2}} \)

---

#### Equation 11: \( \frac{2}{y} + \frac{5}{2} = 4 - \frac{2}{3y} \)

1. Eliminate the fractions by finding a common denominator, which is \( 6y \):
\[
\frac{2}{y} = \frac{12}{6y}, \quad \frac{5}{2} = \frac{15y}{6y}, \quad \frac{2}{3y} = \frac{4}{6y}
\]
So the equation becomes:
\[
\frac{12}{6y} + \frac{15y}{6y} = 4 - \frac{4}{6y}
\]
2. Combine the fractions on the left-hand side:
\[
\frac{12 + 15y}{6y} = 4 - \frac{4}{6y}
\]
3. Eliminate the fractions by multiplying through by \( 6y \):
\[
12 + 15y = 24y - 4
\]
4. Subtract \( 15y \) from both sides:
\[
12 = 9y - 4
\]
5. Add 4 to both sides:
\[
16 = 9y
\]
6. Divide by 9:
\[
y = \frac{16}{9}
\]

Solution: \( \boxed{y = \frac{16}{9}} \)

---

#### Equation 12: \( \frac{3 + x}{6x} = \frac{1}{2x} + \frac{4}{3} \)

1. Eliminate the fractions by finding a common denominator, which is \( 6x \):
\[
\frac{1}{2x} = \frac{3}{6x}, \quad \frac{4}{3} = \frac{8x}{6x}
\]
So the equation becomes:
\[
\frac{3 + x}{6x} = \frac{3}{6x} + \frac{8x}{6x}
\]
2. Combine the fractions on the right-hand side:
\[
\frac{3 + x}{6x} = \frac{3 + 8x}{6x}
\]
3. Since the denominators are the same, equate the numerators:
\[
3 + x = 3 + 8x
\]
4. Subtract \( x \) from both sides:
\[
3 = 3 + 7x
\]
5. Subtract 3 from both sides:
\[
0 = 7x
\]
6. Divide by 7:
\[
x = 0
\]
However, \( x = 0 \) makes the original denominators undefined. Thus, there is no solution.

Solution: \( \boxed{\text{No solution}} \)

---

#### Equation 13: \( \frac{z}{z - 1} = \frac{1}{z - 1} + 2 \)

1. Eliminate the fractions by multiplying through by \( z - 1 \):
\[
z = 1 + 2(z - 1)
\]
2. Distribute the 2 on the right-hand side:
\[
z = 1 + 2z - 2
\]
\[
z = 2z - 1
\]
3. Subtract \( z \) from both sides:
\[
0 = z - 1
\]
4. Add 1 to both sides:
\[
z = 1
\]
However, \( z = 1 \) makes the original denominators undefined. Thus, there is no solution.

Solution: \( \boxed{\text{No solution}} \)

---

#### Equation 14: \( \frac{t}{t - 1} = \frac{2}{t - 1} + 2 \)

1. Eliminate the fractions by multiplying through by \( t - 1 \):
\[
t = 2 + 2(t - 1)
\]
2. Distribute the 2 on the right-hand side:
\[
t = 2 + 2t - 2
\]
\[
t = 2t
\]
3. Subtract \( t \) from both sides:
\[
0 = t
\]
However, \( t = 0 \) does not make the original denominators undefined. Thus, the solution is valid.

Solution: \( \boxed{t = 0} \)

---

#### Equation 15: \( 1 - \frac{x - 3}{x - 2} = \frac{2x - 3}{x - 2} \)

1. Eliminate the fractions by multiplying through by \( x - 2 \):
\[
(x - 2) \left( 1 - \frac{x - 3}{x - 2} \right) = (x - 2) \cdot \frac{2x - 3}{x - 2}
\]
\[
(x - 2) \cdot 1 - (x - 2) \cdot \frac{x - 3}{x - 2} = 2x - 3
\]
\[
x - 2 - (x - 3) = 2x - 3
\]
2. Simplify the left-hand side:
\[
x - 2 - x + 3 = 2x - 3
\]
\[
1 = 2x - 3
\]
3. Add 3 to both sides:
\[
4 = 2x
\]
4. Divide by 2:
\[
x = 2
\]
However, \( x = 2 \) makes the original denominators undefined. Thus, there is no solution.

Solution: \( \boxed{\text{No solution}} \)

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Final Answers:

1. \( \boxed{a = 0} \)
2. \( \boxed{p = 1} \)
3. \( \boxed{x = 3} \)
4. \( \boxed{n = -1} \)
5. \( \boxed{\text{No solution}} \)
6. \( \boxed{x = 4} \)
7. \( \boxed{a = \frac{31}{16}} \)
8. \( \boxed{x = \frac{16}{5}} \)
9. \( \boxed{x = \frac{35}{4}} \)
10. \( \boxed{m = \frac{5}{2}} \)
11. \( \boxed{y = \frac{16}{9}} \)
12. \( \boxed{\text{No solution}} \)
13. \( \boxed{\text{No solution}} \)
14. \( \boxed{t = 0} \)
15. \( \boxed{\text{No solution}} \)