Class 8 Linear Equation of One Variable worksheet with fill-in-the-blank problems.

Problem: Solve the linear equations to find the value of the variables.

We will solve each equation step by step. Let's go through them one by one.

---

#### (1) Solve the equations to find the value of the variables

##### A)
\[
-4(4y - 5) + 3(-5y + 4) + 5(y + 3) = -83
\]

1. Expand each term:
\[
-4(4y - 5) = -16y + 20
\]
\[
3(-5y + 4) = -15y + 12
\]
\[
5(y + 3) = 5y + 15
\]

2. Substitute back into the equation:
\[
-16y + 20 - 15y + 12 + 5y + 15 = -83
\]

3. Combine like terms:
\[
(-16y - 15y + 5y) + (20 + 12 + 15) = -83
\]
\[
-26y + 47 = -83
\]

4. Isolate \( y \):
\[
-26y = -83 - 47
\]
\[
-26y = -130
\]
\[
y = \frac{-130}{-26}
\]
\[
y = 5
\]

Answer:
\[
\boxed{5}
\]

##### B)
\[
4(3b - 1) - 2(-5b + 2) = 36
\]

1. Expand each term:
\[
4(3b - 1) = 12b - 4
\]
\[
-2(-5b + 2) = 10b - 4
\]

2. Substitute back into the equation:
\[
12b - 4 + 10b - 4 = 36
\]

3. Combine like terms:
\[
(12b + 10b) + (-4 - 4) = 36
\]
\[
22b - 8 = 36
\]

4. Isolate \( b \):
\[
22b = 36 + 8
\]
\[
22b = 44
\]
\[
b = \frac{44}{22}
\]
\[
b = 2
\]

Answer:
\[
\boxed{2}
\]

##### C)
\[
5(-3a - 1) + 2(-5a - 2) = -34
\]

1. Expand each term:
\[
5(-3a - 1) = -15a - 5
\]
\[
2(-5a - 2) = -10a - 4
\]

2. Substitute back into the equation:
\[
-15a - 5 - 10a - 4 = -34
\]

3. Combine like terms:
\[
(-15a - 10a) + (-5 - 4) = -34
\]
\[
-25a - 9 = -34
\]

4. Isolate \( a \):
\[
-25a = -34 + 9
\]
\[
-25a = -25
\]
\[
a = \frac{-25}{-25}
\]
\[
a = 1
\]

Answer:
\[
\boxed{1}
\]

##### D)
\[
-4(2b - 4) + 2(2b + 1) + 3(-b + 3) = 20
\]

1. Expand each term:
\[
-4(2b - 4) = -8b + 16
\]
\[
2(2b + 1) = 4b + 2
\]
\[
3(-b + 3) = -3b + 9
\]

2. Substitute back into the equation:
\[
-8b + 16 + 4b + 2 - 3b + 9 = 20
\]

3. Combine like terms:
\[
(-8b + 4b - 3b) + (16 + 2 + 9) = 20
\]
\[
-7b + 27 = 20
\]

4. Isolate \( b \):
\[
-7b = 20 - 27
\]
\[
-7b = -7
\]
\[
b = \frac{-7}{-7}
\]
\[
b = 1
\]

Answer:
\[
\boxed{1}
\]

##### E)
\[
-2(4y - 2) - 2(3y + 1) = 30
\]

1. Expand each term:
\[
-2(4y - 2) = -8y + 4
\]
\[
-2(3y + 1) = -6y - 2
\]

2. Substitute back into the equation:
\[
-8y + 4 - 6y - 2 = 30
\]

3. Combine like terms:
\[
(-8y - 6y) + (4 - 2) = 30
\]
\[
-14y + 2 = 30
\]

4. Isolate \( y \):
\[
-14y = 30 - 2
\]
\[
-14y = 28
\]
\[
y = \frac{28}{-14}
\]
\[
y = -2
\]

Answer:
\[
\boxed{-2}
\]

---

#### (2) Solve the equations to find the value of the variables

##### A)
\[
-3(5y - 5) + 4(-5y - 4) - 3(-4y - 1) + 4(5y - 5) = -18
\]

1. Expand each term:
\[
-3(5y - 5) = -15y + 15
\]
\[
4(-5y - 4) = -20y - 16
\]
\[
-3(-4y - 1) = 12y + 3
\]
\[
4(5y - 5) = 20y - 20
\]

2. Substitute back into the equation:
\[
-15y + 15 - 20y - 16 + 12y + 3 + 20y - 20 = -18
\]

3. Combine like terms:
\[
(-15y - 20y + 12y + 20y) + (15 - 16 + 3 - 20) = -18
\]
\[
7y - 18 = -18
\]

4. Isolate \( y \):
\[
7y = -18 + 18
\]
\[
7y = 0
\]
\[
y = 0
\]

Answer:
\[
\boxed{0}
\]

##### B)
\[
4(2a + 5) - 3(3a + 5) + 2(-a - 1) = 15
\]

1. Expand each term:
\[
4(2a + 5) = 8a + 20
\]
\[
-3(3a + 5) = -9a - 15
\]
\[
2(-a - 1) = -2a - 2
\]

2. Substitute back into the equation:
\[
8a + 20 - 9a - 15 - 2a - 2 = 15
\]

3. Combine like terms:
\[
(8a - 9a - 2a) + (20 - 15 - 2) = 15
\]
\[
-3a + 3 = 15
\]

4. Isolate \( a \):
\[
-3a = 15 - 3
\]
\[
-3a = 12
\]
\[
a = \frac{12}{-3}
\]
\[
a = -4
\]

Answer:
\[
\boxed{-4}
\]

---

#### (3) Solve the equations to find the value of the variables

##### A)
\[
3(2y + 4) + 4(y + 1) = 26
\]

1. Expand each term:
\[
3(2y + 4) = 6y + 12
\]
\[
4(y + 1) = 4y + 4
\]

2. Substitute back into the equation:
\[
6y + 12 + 4y + 4 = 26
\]

3. Combine like terms:
\[
(6y + 4y) + (12 + 4) = 26
\]
\[
10y + 16 = 26
\]

4. Isolate \( y \):
\[
10y = 26 - 16
\]
\[
10y = 10
\]
\[
y = \frac{10}{10}
\]
\[
y = 1
\]

Answer:
\[
\boxed{1}
\]

##### B)
\[
-5(4x + 1) - 3(3x - 5) = -48
\]

1. Expand each term:
\[
-5(4x + 1) = -20x - 5
\]
\[
-3(3x - 5) = -9x + 15
\]

2. Substitute back into the equation:
\[
-20x - 5 - 9x + 15 = -48
\]

3. Combine like terms:
\[
(-20x - 9x) + (-5 + 15) = -48
\]
\[
-29x + 10 = -48
\]

4. Isolate \( x \):
\[
-29x = -48 - 10
\]
\[
-29x = -58
\]
\[
x = \frac{-58}{-29}
\]
\[
x = 2
\]

Answer:
\[
\boxed{2}
\]

---

#### (4) Solve the equations to find the value of the variables

##### A)
\[
-1(5b - 5) - 1(-2b - 1) = -3
\]

1. Expand each term:
\[
-1(5b - 5) = -5b + 5
\]
\[
-1(-2b - 1) = 2b + 1
\]

2. Substitute back into the equation:
\[
-5b + 5 + 2b + 1 = -3
\]

3. Combine like terms:
\[
(-5b + 2b) + (5 + 1) = -3
\]
\[
-3b + 6 = -3
\]

4. Isolate \( b \):
\[
-3b = -3 - 6
\]
\[
-3b = -9
\]
\[
b = \frac{-9}{-3}
\]
\[
b = 3
\]

Answer:
\[
\boxed{3}
\]

##### B)
\[
4(-2z + 1) - 4(z - 1) = 44
\]

1. Expand each term:
\[
4(-2z + 1) = -8z + 4
\]
\[
-4(z - 1) = -4z + 4
\]

2. Substitute back into the equation:
\[
-8z + 4 - 4z + 4 = 44
\]

3. Combine like terms:
\[
(-8z - 4z) + (4 + 4) = 44
\]
\[
-12z + 8 = 44
\]

4. Isolate \( z \):
\[
-12z = 44 - 8
\]
\[
-12z = 36
\]
\[
z = \frac{36}{-12}
\]
\[
z = -3
\]

Answer:
\[
\boxed{-3}
\]

---

#### (5) Solve the equations to find the value of the variables

##### A)
\[
-5(5a - 3) + 2(-2a - 4) = -138
\]

1. Expand each term:
\[
-5(5a - 3) = -25a + 15
\]
\[
2(-2a - 4) = -4a - 8
\]

2. Substitute back into the equation:
\[
-25a + 15 - 4a - 8 = -138
\]

3. Combine like terms:
\[
(-25a - 4a) + (15 - 8) = -138
\]
\[
-29a + 7 = -138
\]

4. Isolate \( a \):
\[
-29a = -138 - 7
\]
\[
-29a = -145
\]
\[
a = \frac{-145}{-29}
\]
\[
a = 5
\]

Answer:
\[
\boxed{5}
\]

##### B)
\[
3(4y + 4) + 5(-y - 5) - 5(-3y + 3) - 5(-2y - 5) = -3
\]

1. Expand each term:
\[
3(4y + 4) = 12y + 12
\]
\[
5(-y - 5) = -5y - 25
\]
\[
-5(-3y + 3) = 15y - 15
\]
\[
-5(-2y - 5) = 10y + 25
\]

2. Substitute back into the equation:
\[
12y + 12 - 5y - 25 + 15y - 15 + 10y + 25 = -3
\]

3. Combine like terms:
\[
(12y - 5y + 15y + 10y) + (12 - 25 - 15 + 25) = -3
\]
\[
32y + 7 = -3
\]

4. Isolate \( y \):
\[
32y = -3 - 7
\]
\[
32y = -10
\]
\[
y = \frac{-10}{32}
\]
\[
y = -\frac{5}{16}
\]

Answer:
\[
\boxed{-\frac{5}{16}}
\]

---

Final Answers:

1. A) \( \boxed{5} \)
B) \( \boxed{2} \)
C) \( \boxed{1} \)
D) \( \boxed{1} \)
E) \( \boxed{-2} \)

2. A) \( \boxed{0} \)
B) \( \boxed{-4} \)

3. A) \( \boxed{1} \)
B) \( \boxed{2} \)

4. A) \( \boxed{3} \)
B) \( \boxed{-3} \)

5. A) \( \boxed{5} \)
B) \( \boxed{-\frac{5}{16}} \)