Solving Basic Equations Worksheets - Free Printable

Problem: Solve the following equations for the variable \( y \) using inverse operations. Show your work.

#### Equation 1:
\[
(y^2 + 9) - 6 \times (4 + y) = -23
\]

Step 1: Expand the equation.
\[
y^2 + 9 - 6(4 + y) = -23
\]
\[
y^2 + 9 - 24 - 6y = -23
\]

Step 2: Simplify the terms.
\[
y^2 - 6y + 9 - 24 = -23
\]
\[
y^2 - 6y - 15 = -23
\]

Step 3: Move all terms to one side to set the equation to zero.
\[
y^2 - 6y - 15 + 23 = 0
\]
\[
y^2 - 6y + 8 = 0
\]

Step 4: Factor the quadratic equation.
\[
y^2 - 6y + 8 = (y - 4)(y - 2) = 0
\]

Step 5: Solve for \( y \).
\[
y - 4 = 0 \quad \text{or} \quad y - 2 = 0
\]
\[
y = 4 \quad \text{or} \quad y = 2
\]

Solution for Equation 1:
\[
\boxed{y = 4 \text{ or } y = 2}
\]

---

#### Equation 2:
\[
(y^2 + 4) - 12 \times (12 + y) = -151
\]

Step 1: Expand the equation.
\[
y^2 + 4 - 12(12 + y) = -151
\]
\[
y^2 + 4 - 144 - 12y = -151
\]

Step 2: Simplify the terms.
\[
y^2 - 12y + 4 - 144 = -151
\]
\[
y^2 - 12y - 140 = -151
\]

Step 3: Move all terms to one side to set the equation to zero.
\[
y^2 - 12y - 140 + 151 = 0
\]
\[
y^2 - 12y + 11 = 0
\]

Step 4: Factor the quadratic equation.
\[
y^2 - 12y + 11 = (y - 11)(y - 1) = 0
\]

Step 5: Solve for \( y \).
\[
y - 11 = 0 \quad \text{or} \quad y - 1 = 0
\]
\[
y = 11 \quad \text{or} \quad y = 1
\]

Solution for Equation 2:
\[
\boxed{y = 11 \text{ or } y = 1}
\]

---

#### Equation 3:
\[
(y^2 + 4) - 12 \times (2 + y) = -40
\]

Step 1: Expand the equation.
\[
y^2 + 4 - 12(2 + y) = -40
\]
\[
y^2 + 4 - 24 - 12y = -40
\]

Step 2: Simplify the terms.
\[
y^2 - 12y + 4 - 24 = -40
\]
\[
y^2 - 12y - 20 = -40
\]

Step 3: Move all terms to one side to set the equation to zero.
\[
y^2 - 12y - 20 + 40 = 0
\]
\[
y^2 - 12y + 20 = 0
\]

Step 4: Factor the quadratic equation.
\[
y^2 - 12y + 20 = (y - 10)(y - 2) = 0
\]

Step 5: Solve for \( y \).
\[
y - 10 = 0 \quad \text{or} \quad y - 2 = 0
\]
\[
y = 10 \quad \text{or} \quad y = 2
\]

Solution for Equation 3:
\[
\boxed{y = 10 \text{ or } y = 2}
\]

---

#### Equation 4:
\[
(y^2 + 10) - 4 \times (10 + y) = 66
\]

Step 1: Expand the equation.
\[
y^2 + 10 - 4(10 + y) = 66
\]
\[
y^2 + 10 - 40 - 4y = 66
\]

Step 2: Simplify the terms.
\[
y^2 - 4y + 10 - 40 = 66
\]
\[
y^2 - 4y - 30 = 66
\]

Step 3: Move all terms to one side to set the equation to zero.
\[
y^2 - 4y - 30 - 66 = 0
\]
\[
y^2 - 4y - 96 = 0
\]

Step 4: Factor the quadratic equation.
\[
y^2 - 4y - 96 = (y - 12)(y + 8) = 0
\]

Step 5: Solve for \( y \).
\[
y - 12 = 0 \quad \text{or} \quad y + 8 = 0
\]
\[
y = 12 \quad \text{or} \quad y = -8
\]

Solution for Equation 4:
\[
\boxed{y = 12 \text{ or } y = -8}
\]

---

#### Equation 5:
\[
(y^2 + 7) - 3 \times (2 + y) = 109
\]

Step 1: Expand the equation.
\[
y^2 + 7 - 3(2 + y) = 109
\]
\[
y^2 + 7 - 6 - 3y = 109
\]

Step 2: Simplify the terms.
\[
y^2 - 3y + 7 - 6 = 109
\]
\[
y^2 - 3y + 1 = 109
\]

Step 3: Move all terms to one side to set the equation to zero.
\[
y^2 - 3y + 1 - 109 = 0
\]
\[
y^2 - 3y - 108 = 0
\]

Step 4: Factor the quadratic equation.
\[
y^2 - 3y - 108 = (y - 12)(y + 9) = 0
\]

Step 5: Solve for \( y \).
\[
y - 12 = 0 \quad \text{or} \quad y + 9 = 0
\]
\[
y = 12 \quad \text{or} \quad y = -9
\]

Solution for Equation 5:
\[
\boxed{y = 12 \text{ or } y = -9}
\]

---

Final Answers:

1. \(\boxed{y = 4 \text{ or } y = 2}\)
2. \(\boxed{y = 11 \text{ or } y = 1}\)
3. \(\boxed{y = 10 \text{ or } y = 2}\)
4. \(\boxed{y = 12 \text{ or } y = -8}\)
5. \(\boxed{y = 12 \text{ or } y = -9}\)