Practice worksheet for 8th graders to master linear equations by solving for unknown variables.

Explanation:
We are solving linear equations with one variable. The goal is to simplify both sides, collect like terms, and isolate the variable.

Let’s solve each part carefully.

---

(1) A)
Equation:
\[
-4(4y - 5) + 3(-5y + 4) + 5(y + 3) = -83
\]

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

Now add them:
\[
(-16y -15y + 5y) + (20 + 12 + 15) = -83
\]
\[
(-26y) + 47 = -83
\]

Step 2: Move constant to right:
\[
-26y = -83 - 47 = -130
\]

Step 3: Divide:
\[
y = \frac{-130}{-26} = 5
\]

✔ So, y = 5

---

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

Expand:
- \(4(3b - 1) = 12b - 4\)
- \(-2(-5b + 2) = 10b - 4\)

Add:
\[
12b + 10b - 4 - 4 = 36 \Rightarrow 22b - 8 = 36
\]

Add 8:
\[
22b = 44 \Rightarrow b = 2
\]

✔ b = 2

---

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

Expand:
- \(5(-3a - 1) = -15a - 5\)
- \(2(-5a - 2) = -10a - 4\)

Sum:
\[
-15a - 10a - 5 - 4 = -34 \Rightarrow -25a - 9 = -34
\]

Add 9:
\[
-25a = -25 \Rightarrow a = 1
\]

✔ a = 1

---

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

Expand:
- \(-4(2b - 4) = -8b + 16\)
- \(2(2b + 1) = 4b + 2\)
- \(3(-b + 3) = -3b + 9\)

Add all:
\[
(-8b + 4b - 3b) + (16 + 2 + 9) = 20
\Rightarrow (-7b) + 27 = 20
\]

Subtract 27:
\[
-7b = -7 \Rightarrow b = 1
\]

✔ b = 1

---

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

Expand:
- \(-2(4y - 2) = -8y + 4\)
- \(-2(3y + 1) = -6y - 2\)

Sum:
\[
-8y - 6y + 4 - 2 = 30 \Rightarrow -14y + 2 = 30
\]

Subtract 2:
\[
-14y = 28 \Rightarrow y = -2
\]

✔ y = -2

---

(2) A)
Equation:
\[
-3(5y - 5) + 4(-5y - 4) - 3(-4y - 1) + 4(5y - 5) = -18
\]

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

Now combine like terms:

y-terms: \(-15y -20y + 12y + 20y = (-15 -20 +12 +20)y = (-3)y = -3y\)

Constants: \(15 -16 + 3 -20 = (15 + 3) - (16 + 20) = 18 - 36 = -18\)

So left side = \(-3y - 18\)

Equation:
\[
-3y - 18 = -18 \Rightarrow -3y = 0 \Rightarrow y = 0
\]

✔ y = 0

---

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

Expand:
- \(4(2a + 5) = 8a + 20\)
- \(-3(3a + 5) = -9a - 15\)
- \(2(-a - 1) = -2a - 2\)

Combine:
a-terms: \(8a - 9a - 2a = -3a\)
constants: \(20 - 15 - 2 = 3\)

So:
\[
-3a + 3 = 15 \Rightarrow -3a = 12 \Rightarrow a = -4
\]

✔ a = -4

---

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

Expand:
- \(3(2y + 4) = 6y + 12\)
- \(4(y + 1) = 4y + 4\)

Sum:
\(6y + 4y + 12 + 4 = 10y + 16 = 26\)

Then:
\(10y = 10 \Rightarrow y = 1\)

✔ y = 1

---

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

Expand:
- \(-5(4x + 1) = -20x - 5\)
- \(-3(3x - 5) = -9x + 15\)

Sum:
\(-20x - 9x = -29x\), constants: \(-5 + 15 = 10\)

So:
\(-29x + 10 = -48\)
→ \(-29x = -58\)
→ \(x = 2\)

✔ x = 2

---

(4) A)
Equation:
\[
-1(5b - 5) - 1(-2b - 1) = -3
\]

Simplify:
- \(-1(5b - 5) = -5b + 5\)
- \(-1(-2b - 1) = 2b + 1\)

Sum:
\(-5b + 2b + 5 + 1 = -3b + 6 = -3\)

Then:
\(-3b = -9 \Rightarrow b = 3\)

✔ b = 3

---

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

Expand:
- \(4(-2z + 1) = -8z + 4\)
- \(-4(z - 1) = -4z + 4\)

Sum:
\(-8z - 4z + 4 + 4 = -12z + 8 = 44\)

Then:
\(-12z = 36 \Rightarrow z = -3\)

✔ z = -3

---

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

Expand:
- \(-5(5a - 3) = -25a + 15\)
- \(2(-2a - 4) = -4a - 8\)

Sum:
\(-25a - 4a = -29a\), constants: \(15 - 8 = 7\)

So:
\(-29a + 7 = -138\)
→ \(-29a = -145\)
→ \(a = 5\)

✔ a = 5

---

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

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

Now sum y-terms:
\(12y - 5y + 15y + 10y = (12 - 5 + 15 + 10)y = 32y\)

Constants:
\(12 - 25 - 15 + 25 = (12 + 25) - (25 + 15) = 37 - 40 = -3\)

So left side = \(32y - 3\)

Equation:
\(32y - 3 = -3 \Rightarrow 32y = 0 \Rightarrow y = 0\)

✔ y = 0

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(6) — Wait! The problem says “(6) Solve the equations to find the value of the variables”, but no equation is listed under (6) in the image. Since it's missing, we assume it's a typo or incomplete. We’ll skip (6) — only solve what's given.

Now compile all answers:

(1) A) y = 5
(1) B) b = 2
(1) C) a = 1
(1) D) b = 1
(1) E) y = -2

(2) A) y = 0
(2) B) a = -4

(3) A) y = 1
(3) B) x = 2

(4) A) b = 3
(4) B) z = -3

(5) A) a = 5
(5) B) y = 0

Final Answer:
(1) A) 5
(1) B) 2
(1) C) 1
(1) D) 1
(1) E) -2
(2) A) 0
(2) B) -4
(3) A) 1
(3) B) 2
(4) A) 3
(4) B) -3
(5) A) 5
(5) B) 0