Grade 10 Algebra Worksheet - Solving Linear Equations

Problem Analysis:

The task involves solving a series of linear equations, some of which require expanding expressions and dealing with algebraic fractions. The goal is to solve each equation for the unknown variable.

Solution Approach:

1. Expand any brackets if present.
2. Simplify both sides of the equation.
3. Isolate the variable by performing algebraic operations (e.g., addition, subtraction, multiplication, division).
4. Solve for the variable and verify the solution if necessary.

Detailed Solutions:

#### Equations 1 to 10:

1. Equation: \( 4x(2 + 3) = 9(3x - 5) \)
- Expand both sides:
\[
4x \cdot 5 = 9(3x - 5)
\]
\[
20x = 27x - 45
\]
- Rearrange to isolate \( x \):
\[
20x - 27x = -45
\]
\[
-7x = -45
\]
\[
x = \frac{45}{7}
\]

2. Equation: \( 3(x + 4) = 5(x - 6) + 32 \)
- Expand both sides:
\[
3x + 12 = 5x - 30 + 32
\]
\[
3x + 12 = 5x + 2
\]
- Rearrange to isolate \( x \):
\[
3x - 5x = 2 - 12
\]
\[
-2x = -10
\]
\[
x = 5
\]

3. Equation: \( 2(x + 3) - 4(x - 1) = 0 \)
- Expand both terms:
\[
2x + 6 - 4x + 4 = 0
\]
\[
-2x + 10 = 0
\]
- Solve for \( x \):
\[
-2x = -10
\]
\[
x = 5
\]

4. Equation: \( 4x - 2(x - 1) = 1 \)
- Expand and simplify:
\[
4x - 2x + 2 = 1
\]
\[
2x + 2 = 1
\]
- Solve for \( x \):
\[
2x = 1 - 2
\]
\[
2x = -1
\]
\[
x = -\frac{1}{2}
\]

5. Equation: \( 7(x + 3) = 9(2x - 1) - 3 \)
- Expand both sides:
\[
7x + 21 = 18x - 9 - 3
\]
\[
7x + 21 = 18x - 12
\]
- Rearrange to isolate \( x \):
\[
7x - 18x = -12 - 21
\]
\[
-11x = -33
\]
\[
x = 3
\]

6. Equation: \( 6(x + 1) = 6x \)
- Expand the left side:
\[
6x + 6 = 6x
\]
- Subtract \( 6x \) from both sides:
\[
6 = 0
\]
- This is a contradiction, so there is no solution.

7. Equation: \( 7x - 5(x - 2) = 12 \)
- Expand and simplify:
\[
7x - 5x + 10 = 12
\]
\[
2x + 10 = 12
\]
- Solve for \( x \):
\[
2x = 12 - 10
\]
\[
2x = 2
\]
\[
x = 1
\]

8. Equation: \( 4x - (x + 3) = 2 - (3 - x) \)
- Expand both sides:
\[
4x - x - 3 = 2 - 3 + x
\]
\[
3x - 3 = -1 + x
\]
- Rearrange to isolate \( x \):
\[
3x - x = -1 + 3
\]
\[
2x = 2
\]
\[
x = 1
\]

9. Equation: \( 5(x + 2) - 3 = 2x - (x - 1) \)
- Expand both sides:
\[
5x + 10 - 3 = 2x - x + 1
\]
\[
5x + 7 = x + 1
\]
- Rearrange to isolate \( x \):
\[
5x - x = 1 - 7
\]
\[
4x = -6
\]
\[
x = -\frac{3}{2}
\]

10. Equation: \( 3(x - 2) = 2(x + 3) \)
- Expand both sides:
\[
3x - 6 = 2x + 6
\]
- Rearrange to isolate \( x \):
\[
3x - 2x = 6 + 6
\]
\[
x = 12
\]

#### Equations 11 to 15:

11. Equation: \( \frac{2x - 3}{7} = \frac{3x - 5}{10} \)
- Cross-multiply:
\[
10(2x - 3) = 7(3x - 5)
\]
\[
20x - 30 = 21x - 35
\]
- Rearrange to isolate \( x \):
\[
20x - 21x = -35 + 30
\]
\[
-x = -5
\]
\[
x = 5
\]

12. Equation: \( \frac{2x}{3} = \frac{x + 1}{4} \)
- Cross-multiply:
\[
4(2x) = 3(x + 1)
\]
\[
8x = 3x + 3
\]
- Rearrange to isolate \( x \):
\[
8x - 3x = 3
\]
\[
5x = 3
\]
\[
x = \frac{3}{5}
\]

13. Equation: \( \frac{3x + 1}{5} = \frac{2x - 1}{3} \)
- Cross-multiply:
\[
3(3x + 1) = 5(2x - 1)
\]
\[
9x + 3 = 10x - 5
\]
- Rearrange to isolate \( x \):
\[
9x - 10x = -5 - 3
\]
\[
-x = -8
\]
\[
x = 8
\]

14. Equation: \( \frac{5x - 2}{4} + \frac{3x + 2}{2} = 2x \)
- Find a common denominator (4):
\[
\frac{5x - 2}{4} + \frac{2(3x + 2)}{4} = 2x
\]
\[
\frac{5x - 2 + 6x + 4}{4} = 2x
\]
\[
\frac{11x + 2}{4} = 2x
\]
- Multiply through by 4:
\[
11x + 2 = 8x
\]
- Rearrange to isolate \( x \):
\[
11x - 8x = -2
\]
\[
3x = -2
\]
\[
x = -\frac{2}{3}
\]

15. Equation: \( \frac{5m + 3}{5} = \frac{m + 2}{3} \)
- Cross-multiply:
\[
3(5m + 3) = 5(m + 2)
\]
\[
15m + 9 = 5m + 10
\]
- Rearrange to isolate \( m \):
\[
15m - 5m = 10 - 9
\]
\[
10m = 1
\]
\[
m = \frac{1}{10}
\]

Final Answers:

\[
\boxed{
\begin{aligned}
1. & \ x = \frac{45}{7} \\
2. & \ x = 5 \\
3. & \ x = 5 \\
4. & \ x = -\frac{1}{2} \\
5. & \ x = 3 \\
6. & \ \text{No solution} \\
7. & \ x = 1 \\
8. & \ x = 1 \\
9. & \ x = -\frac{3}{2} \\
10. & \ x = 12 \\
11. & \ x = 5 \\
12. & \ x = \frac{3}{5} \\
13. & \ x = 8 \\
14. & \ x = -\frac{2}{3} \\
15. & \ m = \frac{1}{10}
\end{aligned}
}
\]