Seventh Grade Solving Equations With Parentheses - Twinkl - Free Printable

Problem: Solving Equations with Parentheses

The task involves solving each of the given equations step by step. Below, I will solve all six equations systematically and explain each step.

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#### Equation 1: \( 5(z - 7) = 10 \)

1. Distribute the 5 inside the parentheses:
\[
5(z - 7) = 5z - 35
\]
So the equation becomes:
\[
5z - 35 = 10
\]

2. Add 35 to both sides to isolate the term with \( z \):
\[
5z - 35 + 35 = 10 + 35
\]
Simplifying:
\[
5z = 45
\]

3. Divide both sides by 5 to solve for \( z \):
\[
\frac{5z}{5} = \frac{45}{5}
\]
Simplifying:
\[
z = 9
\]

Solution for Equation 1:
\[
\boxed{z = 9}
\]

---

#### Equation 2: \( 30 = 6(z - 3) \)

1. Distribute the 6 inside the parentheses:
\[
6(z - 3) = 6z - 18
\]
So the equation becomes:
\[
30 = 6z - 18
\]

2. Add 18 to both sides to isolate the term with \( z \):
\[
30 + 18 = 6z - 18 + 18
\]
Simplifying:
\[
48 = 6z
\]

3. Divide both sides by 6 to solve for \( z \):
\[
\frac{48}{6} = \frac{6z}{6}
\]
Simplifying:
\[
z = 8
\]

Solution for Equation 2:
\[
\boxed{z = 8}
\]

---

#### Equation 3: \( -4(z - 8) = 56 \)

1. Distribute the \(-4\) inside the parentheses:
\[
-4(z - 8) = -4z + 32
\]
So the equation becomes:
\[
-4z + 32 = 56
\]

2. Subtract 32 from both sides to isolate the term with \( z \):
\[
-4z + 32 - 32 = 56 - 32
\]
Simplifying:
\[
-4z = 24
\]

3. Divide both sides by \(-4\) to solve for \( z \):
\[
\frac{-4z}{-4} = \frac{24}{-4}
\]
Simplifying:
\[
z = -6
\]

Solution for Equation 3:
\[
\boxed{z = -6}
\]

---

#### Equation 4: \( 44 = 2(k + 15) \)

1. Distribute the 2 inside the parentheses:
\[
2(k + 15) = 2k + 30
\]
So the equation becomes:
\[
44 = 2k + 30
\]

2. Subtract 30 from both sides to isolate the term with \( k \):
\[
44 - 30 = 2k + 30 - 30
\]
Simplifying:
\[
14 = 2k
\]

3. Divide both sides by 2 to solve for \( k \):
\[
\frac{14}{2} = \frac{2k}{2}
\]
Simplifying:
\[
k = 7
\]

Solution for Equation 4:
\[
\boxed{k = 7}
\]

---

#### Equation 5: \( 103 = 9(z - 4) \)

1. Distribute the 9 inside the parentheses:
\[
9(z - 4) = 9z - 36
\]
So the equation becomes:
\[
103 = 9z - 36
\]

2. Add 36 to both sides to isolate the term with \( z \):
\[
103 + 36 = 9z - 36 + 36
\]
Simplifying:
\[
139 = 9z
\]

3. Divide both sides by 9 to solve for \( z \):
\[
\frac{139}{9} = \frac{9z}{9}
\]
Simplifying:
\[
z = \frac{139}{9}
\]

Solution for Equation 5:
\[
\boxed{z = \frac{139}{9}}
\]

---

#### Equation 6: \( 7(z + 12) = 77 \)

1. Distribute the 7 inside the parentheses:
\[
7(z + 12) = 7z + 84
\]
So the equation becomes:
\[
7z + 84 = 77
\]

2. Subtract 84 from both sides to isolate the term with \( z \):
\[
7z + 84 - 84 = 77 - 84
\]
Simplifying:
\[
7z = -7
\]

3. Divide both sides by 7 to solve for \( z \):
\[
\frac{7z}{7} = \frac{-7}{7}
\]
Simplifying:
\[
z = -1
\]

Solution for Equation 6:
\[
\boxed{z = -1}
\]

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Final Answers:

1. \( z = 9 \)
2. \( z = 8 \)
3. \( z = -6 \)
4. \( k = 7 \)
5. \( z = \frac{139}{9} \)
6. \( z = -1 \)

\[
\boxed{9, 8, -6, 7, \frac{139}{9}, -1}
\]