Linear Equations worksheet featuring five math problems for practice.

Let's solve each question step by step.

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Question 1: Solve the linear equations

#### Part A: \( 8x + 3 = 27 + 2x \)

1. Start with the given equation:
\[
8x + 3 = 27 + 2x
\]

2. Subtract \( 2x \) from both sides to isolate \( x \)-terms on one side:
\[
8x - 2x + 3 = 27
\]
\[
6x + 3 = 27
\]

3. Subtract 3 from both sides to isolate the term with \( x \):
\[
6x = 24
\]

4. Divide both sides by 6 to solve for \( x \):
\[
x = 4
\]

Solution for Part A:
\[
\boxed{x = 4}
\]

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#### Part B: \( \frac{7y}{5} = y - 4 \)

1. Start with the given equation:
\[
\frac{7y}{5} = y - 4
\]

2. Eliminate the fraction by multiplying every term by 5:
\[
5 \cdot \frac{7y}{5} = 5 \cdot (y - 4)
\]
\[
7y = 5y - 20
\]

3. Subtract \( 5y \) from both sides to isolate \( y \)-terms on one side:
\[
7y - 5y = -20
\]
\[
2y = -20
\]

4. Divide both sides by 2 to solve for \( y \):
\[
y = -10
\]

Solution for Part B:
\[
\boxed{y = -10}
\]

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Question 2: Two numbers are such that the ratio between them is 3 : 5. If each is increased by 10, the ratio between the new numbers formed is 5 : 7. Find the original numbers.

1. Let the two numbers be \( 3x \) and \( 5x \), where \( x \) is a common multiplier.

2. When each number is increased by 10, the new numbers become:
\[
3x + 10 \quad \text{and} \quad 5x + 10
\]

3. According to the problem, the new ratio is 5 : 7:
\[
\frac{3x + 10}{5x + 10} = \frac{5}{7}
\]

4. Cross-multiply to solve for \( x \):
\[
7(3x + 10) = 5(5x + 10)
\]
\[
21x + 70 = 25x + 50
\]

5. Subtract \( 21x \) from both sides:
\[
70 = 4x + 50
\]

6. Subtract 50 from both sides:
\[
20 = 4x
\]

7. Divide both sides by 4:
\[
x = 5
\]

8. Substitute \( x = 5 \) back into the expressions for the original numbers:
\[
3x = 3 \cdot 5 = 15
\]
\[
5x = 5 \cdot 5 = 25
\]

Solution for Question 2:
\[
\boxed{15 \text{ and } 25}
\]

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Question 3: Half of a herd of deer are grazing in the field and three-fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

1. Let the total number of deer in the herd be \( D \).

2. Half of the herd is grazing:
\[
\text{Grazing deer} = \frac{D}{2}
\]

3. The remaining deer after grazing:
\[
\text{Remaining deer} = D - \frac{D}{2} = \frac{D}{2}
\]

4. Three-fourths of the remaining deer are playing nearby:
\[
\text{Playing deer} = \frac{3}{4} \cdot \frac{D}{2} = \frac{3D}{8}
\]

5. The rest of the deer (those not grazing or playing) are drinking water:
\[
\text{Drinking deer} = \frac{D}{2} - \frac{3D}{8}
\]

6. To simplify \( \frac{D}{2} - \frac{3D}{8} \):
\[
\frac{D}{2} = \frac{4D}{8}
\]
\[
\frac{D}{2} - \frac{3D}{8} = \frac{4D}{8} - \frac{3D}{8} = \frac{D}{8}
\]

7. According to the problem, the number of deer drinking water is 9:
\[
\frac{D}{8} = 9
\]

8. Solve for \( D \):
\[
D = 9 \cdot 8 = 72
\]

Solution for Question 3:
\[
\boxed{72}
\]

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Question 4: Find the three consecutive even numbers whose sum is 234.

1. Let the three consecutive even numbers be \( x \), \( x + 2 \), and \( x + 4 \).

2. Their sum is given as 234:
\[
x + (x + 2) + (x + 4) = 234
\]

3. Simplify the equation:
\[
x + x + 2 + x + 4 = 234
\]
\[
3x + 6 = 234
\]

4. Subtract 6 from both sides:
\[
3x = 228
\]

5. Divide both sides by 3:
\[
x = 76
\]

6. The three consecutive even numbers are:
\[
x = 76, \quad x + 2 = 78, \quad x + 4 = 80
\]

Solution for Question 4:
\[
\boxed{76, 78, 80}
\]

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Question 5: Mary’s father is 26 years younger than Mary’s grandfather and 29 years older than Mary. The sum of the ages of all three is 135 years. What is the age of each one of them?

1. Let Mary’s age be \( M \).
2. Let Mary’s father’s age be \( F \).
3. Let Mary’s grandfather’s age be \( G \).

4. From the problem, we have the following relationships:
\[
F = G - 26 \quad \text{(Mary’s father is 26 years younger than her grandfather)}
\]
\[
F = M + 29 \quad \text{(Mary’s father is 29 years older than Mary)}
\]
\[
M + F + G = 135 \quad \text{(The sum of their ages is 135)}
\]

5. Substitute \( F = M + 29 \) into \( F = G - 26 \):
\[
M + 29 = G - 26
\]
\[
G = M + 29 + 26
\]
\[
G = M + 55
\]

6. Substitute \( F = M + 29 \) and \( G = M + 55 \) into the sum equation \( M + F + G = 135 \):
\[
M + (M + 29) + (M + 55) = 135
\]
\[
M + M + 29 + M + 55 = 135
\]
\[
3M + 84 = 135
\]

7. Subtract 84 from both sides:
\[
3M = 51
\]

8. Divide both sides by 3:
\[
M = 17
\]

9. Find \( F \) and \( G \):
\[
F = M + 29 = 17 + 29 = 46
\]
\[
G = M + 55 = 17 + 55 = 72
\]

Solution for Question 5:
\[
\boxed{17, 46, 72}
\]

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Final Answers:

1. \( \boxed{x = 4} \) and \( \boxed{y = -10} \)
2. \( \boxed{15 \text{ and } 25} \)
3. \( \boxed{72} \)
4. \( \boxed{76, 78, 80} \)
5. \( \boxed{17, 46, 72} \)