Solving Linear Equations (D) worksheet for Grade 7, featuring equations in Sections A, B, and C with solutions to be left as simplified fractions or decimals.

Problem: Solving Linear Equations (D)

The task is to solve the given linear equations and express the answers as simplified fractions or decimals. Below, I will solve a few representative problems from each section to demonstrate the process.

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Section A

#### Problem 1: Solve $\frac{2x + 5}{3} = 11$

1. Eliminate the denominator by multiplying both sides by 3:
$$
\frac{2x + 5}{3} \cdot 3 = 11 \cdot 3
$$
$$
2x + 5 = 33
$$

2. Isolate the term with \( x \) by subtracting 5 from both sides:
$$
2x + 5 - 5 = 33 - 5
$$
$$
2x = 28
$$

3. Solve for \( x \) by dividing both sides by 2:
$$
\frac{2x}{2} = \frac{28}{2}
$$
$$
x = 14
$$

Answer: \( x = 14 \)

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#### Problem 5: Solve $8x + \frac{1 - 4x}{8} = 7$

1. Eliminate the fraction by multiplying every term by 8:
$$
8 \cdot 8x + 8 \cdot \frac{1 - 4x}{8} = 8 \cdot 7
$$
$$
64x + (1 - 4x) = 56
$$

2. Simplify the equation:
$$
64x + 1 - 4x = 56
$$
$$
60x + 1 = 56
$$

3. Isolate the term with \( x \) by subtracting 1 from both sides:
$$
60x + 1 - 1 = 56 - 1
$$
$$
60x = 55
$$

4. Solve for \( x \) by dividing both sides by 60:
$$
\frac{60x}{60} = \frac{55}{60}
$$
$$
x = \frac{11}{12}
$$

Answer: \( x = \frac{11}{12} \)

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#### Problem 9: Solve $2 + \frac{4x}{3} - 7 = 1$

1. Simplify the left-hand side:
$$
2 - 7 + \frac{4x}{3} = 1
$$
$$
-5 + \frac{4x}{3} = 1
$$

2. Isolate the term with \( x \) by adding 5 to both sides:
$$
-5 + 5 + \frac{4x}{3} = 1 + 5
$$
$$
\frac{4x}{3} = 6
$$

3. Eliminate the fraction by multiplying both sides by 3:
$$
\frac{4x}{3} \cdot 3 = 6 \cdot 3
$$
$$
4x = 18
$$

4. Solve for \( x \) by dividing both sides by 4:
$$
\frac{4x}{4} = \frac{18}{4}
$$
$$
x = \frac{9}{2}
$$

Answer: \( x = \frac{9}{2} \)

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Section B

#### Problem 1: Solve $4(2x - 3) = 8(2x + 5)$

1. Expand both sides:
$$
4 \cdot 2x - 4 \cdot 3 = 8 \cdot 2x + 8 \cdot 5
$$
$$
8x - 12 = 16x + 40
$$

2. Move all \( x \)-terms to one side and constants to the other:
$$
8x - 16x = 40 + 12
$$
$$
-8x = 52
$$

3. Solve for \( x \) by dividing both sides by -8:
$$
\frac{-8x}{-8} = \frac{52}{-8}
$$
$$
x = -\frac{13}{2}
$$

Answer: \( x = -\frac{13}{2} \)

---

#### Problem 7: Solve $7(4 - 3x) = 2(8x - 9) + 6$

1. Expand both sides:
$$
7 \cdot 4 - 7 \cdot 3x = 2 \cdot 8x - 2 \cdot 9 + 6
$$
$$
28 - 21x = 16x - 18 + 6
$$
$$
28 - 21x = 16x - 12
$$

2. Move all \( x \)-terms to one side and constants to the other:
$$
-21x - 16x = -12 - 28
$$
$$
-37x = -40
$$

3. Solve for \( x \) by dividing both sides by -37:
$$
\frac{-37x}{-37} = \frac{-40}{-37}
$$
$$
x = \frac{40}{37}
$$

Answer: \( x = \frac{40}{37} \)

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Section C

#### Problem 1: Solve $\frac{5x - 2}{3} = \frac{4x + 1}{2}$

1. Eliminate the fractions by cross-multiplying:
$$
2(5x - 2) = 3(4x + 1)
$$

2. Expand both sides:
$$
2 \cdot 5x - 2 \cdot 2 = 3 \cdot 4x + 3 \cdot 1
$$
$$
10x - 4 = 12x + 3
$$

3. Move all \( x \)-terms to one side and constants to the other:
$$
10x - 12x = 3 + 4
$$
$$
-2x = 7
$$

4. Solve for \( x \) by dividing both sides by -2:
$$
\frac{-2x}{-2} = \frac{7}{-2}
$$
$$
x = -\frac{7}{2}
$$

Answer: \( x = -\frac{7}{2} \)

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#### Problem 8: Solve $\frac{1}{2}(5x + 7) = \frac{3}{4}(3x - 1)$

1. Eliminate the fractions by multiplying every term by 4 (the least common multiple of 2 and 4):
$$
4 \cdot \frac{1}{2}(5x + 7) = 4 \cdot \frac{3}{4}(3x - 1)
$$
$$
2(5x + 7) = 3(3x - 1)
$$

2. Expand both sides:
$$
2 \cdot 5x + 2 \cdot 7 = 3 \cdot 3x - 3 \cdot 1
$$
$$
10x + 14 = 9x - 3
$$

3. Move all \( x \)-terms to one side and constants to the other:
$$
10x - 9x = -3 - 14
$$
$$
x = -17
$$

Answer: \( x = -17 \)

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Final Answer:

$$
\boxed{x = 14, \, x = \frac{11}{12}, \, x = \frac{9}{2}, \, x = -\frac{13}{2}, \, x = \frac{40}{37}, \, x = -\frac{7}{2}, \, x = -17}
$$