Solving Equations - Rectangles worksheet with six problems involving algebraic expressions for length and width of rectangles.

To solve the problems involving rectangles and their areas, we need to use the formula for the area of a rectangle:

\[
\text{Area} = \text{length} \times \text{width}
\]

We will solve each problem step by step.

---

Problem 1:

\[
\text{Area} = 80 \, \text{ft}^2
\]
\[
\text{Length} = 4x \, \text{ft}, \quad \text{Width} = (x + 8) \, \text{ft}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
80 = 4x \cdot (x + 8)
\]

#### Step 2: Expand the right-hand side.
\[
80 = 4x^2 + 32x
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
4x^2 + 32x - 80 = 0
\]

#### Step 4: Simplify by dividing the entire equation by 4.
\[
x^2 + 8x - 20 = 0
\]

#### Step 5: Solve the quadratic equation using factoring.
We need two numbers that multiply to \(-20\) and add to \(8\). These numbers are \(10\) and \(-2\).
\[
x^2 + 8x - 20 = (x + 10)(x - 2) = 0
\]

#### Step 6: Solve for \(x\).
\[
x + 10 = 0 \quad \text{or} \quad x - 2 = 0
\]
\[
x = -10 \quad \text{or} \quad x = 2
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = 2
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = 4x = 4(2) = 8 \, \text{ft}
\]
\[
\text{Width} = x + 8 = 2 + 8 = 10 \, \text{ft}
\]

#### Final Answer for Problem 1:
\[
x = 2, \quad \text{Length} = 8 \, \text{ft}, \quad \text{Width} = 10 \, \text{ft}
\]

---

Problem 2:

\[
\text{Area} = 600 \, \text{yd}^2
\]
\[
\text{Length} = 5x \, \text{yd}, \quad \text{Width} = (3x + 2) \, \text{yd}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
600 = 5x \cdot (3x + 2)
\]

#### Step 2: Expand the right-hand side.
\[
600 = 15x^2 + 10x
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
15x^2 + 10x - 600 = 0
\]

#### Step 4: Simplify by dividing the entire equation by 5.
\[
3x^2 + 2x - 120 = 0
\]

#### Step 5: Solve the quadratic equation using the quadratic formula.
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 3\), \(b = 2\), and \(c = -120\).

\[
x = \frac{-2 \pm \sqrt{2^2 - 4(3)(-120)}}{2(3)}
\]
\[
x = \frac{-2 \pm \sqrt{4 + 1440}}{6}
\]
\[
x = \frac{-2 \pm \sqrt{1444}}{6}
\]
\[
x = \frac{-2 \pm 38}{6}
\]

#### Step 6: Solve for \(x\).
\[
x = \frac{-2 + 38}{6} = \frac{36}{6} = 6
\]
\[
x = \frac{-2 - 38}{6} = \frac{-40}{6} = -\frac{20}{3}
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = 6
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = 5x = 5(6) = 30 \, \text{yd}
\]
\[
\text{Width} = 3x + 2 = 3(6) + 2 = 18 + 2 = 20 \, \text{yd}
\]

#### Final Answer for Problem 2:
\[
x = 6, \quad \text{Length} = 30 \, \text{yd}, \quad \text{Width} = 20 \, \text{yd}
\]

---

Problem 3:

\[
\text{Area} = 54 \, \text{yd}^2
\]
\[
\text{Length} = (x + 6) \, \text{yd}, \quad \text{Width} = (x - 2) \, \text{yd}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
54 = (x + 6)(x - 2)
\]

#### Step 2: Expand the right-hand side.
\[
54 = x^2 + 6x - 2x - 12
\]
\[
54 = x^2 + 4x - 12
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
x^2 + 4x - 12 - 54 = 0
\]
\[
x^2 + 4x - 66 = 0
\]

#### Step 4: Solve the quadratic equation using the quadratic formula.
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 1\), \(b = 4\), and \(c = -66\).

\[
x = \frac{-4 \pm \sqrt{4^2 - 4(1)(-66)}}{2(1)}
\]
\[
x = \frac{-4 \pm \sqrt{16 + 264}}{2}
\]
\[
x = \frac{-4 \pm \sqrt{280}}{2}
\]
\[
x = \frac{-4 \pm 2\sqrt{70}}{2}
\]
\[
x = -2 \pm \sqrt{70}
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = -2 + \sqrt{70}
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = x + 6 = (-2 + \sqrt{70}) + 6 = 4 + \sqrt{70} \, \text{yd}
\]
\[
\text{Width} = x - 2 = (-2 + \sqrt{70}) - 2 = -4 + \sqrt{70} \, \text{yd}
\]

#### Final Answer for Problem 3:
\[
x = -2 + \sqrt{70}, \quad \text{Length} = 4 + \sqrt{70} \, \text{yd}, \quad \text{Width} = -4 + \sqrt{70} \, \text{yd}
\]

---

Problem 4:

\[
\text{Area} = 28 \, \text{m}^2
\]
\[
\text{Length} = (x + 2) \, \text{m}, \quad \text{Width} = (x - 2) \, \text{m}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
28 = (x + 2)(x - 2)
\]

#### Step 2: Expand the right-hand side.
\[
28 = x^2 - 4
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
x^2 - 4 - 28 = 0
\]
\[
x^2 - 32 = 0
\]

#### Step 4: Solve for \(x\).
\[
x^2 = 32
\]
\[
x = \pm \sqrt{32}
\]
\[
x = \pm 4\sqrt{2}
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = 4\sqrt{2}
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = x + 2 = 4\sqrt{2} + 2 \, \text{m}
\]
\[
\text{Width} = x - 2 = 4\sqrt{2} - 2 \, \text{m}
\]

#### Final Answer for Problem 4:
\[
x = 4\sqrt{2}, \quad \text{Length} = 4\sqrt{2} + 2 \, \text{m}, \quad \text{Width} = 4\sqrt{2} - 2 \, \text{m}
\]

---

Problem 5:

\[
\text{Area} = 120 \, \text{in}^2
\]
\[
\text{Length} = (x + 4) \, \text{in}, \quad \text{Width} = (x - 4) \, \text{in}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
120 = (x + 4)(x - 4)
\]

#### Step 2: Expand the right-hand side.
\[
120 = x^2 - 16
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
x^2 - 16 - 120 = 0
\]
\[
x^2 - 136 = 0
\]

#### Step 4: Solve for \(x\).
\[
x^2 = 136
\]
\[
x = \pm \sqrt{136}
\]
\[
x = \pm 2\sqrt{34}
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = 2\sqrt{34}
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = x + 4 = 2\sqrt{34} + 4 \, \text{in}
\]
\[
\text{Width} = x - 4 = 2\sqrt{34} - 4 \, \text{in}
\]

#### Final Answer for Problem 5:
\[
x = 2\sqrt{34}, \quad \text{Length} = 2\sqrt{34} + 4 \, \text{in}, \quad \text{Width} = 2\sqrt{34} - 4 \, \text{in}
\]

---

Problem 6:

\[
\text{Area} = 40 \, \text{ft}^2
\]
\[
\text{Length} = (5x - 4) \, \text{ft}, \quad \text{Width} = (x + 2) \, \text{ft}
\]

#### Step 1: Set up the equation using the area formula.
\[
\text{Area} = \text{length} \times \text{width}
\]
\[
40 = (5x - 4)(x + 2)
\]

#### Step 2: Expand the right-hand side.
\[
40 = 5x^2 + 10x - 4x - 8
\]
\[
40 = 5x^2 + 6x - 8
\]

#### Step 3: Rearrange the equation to standard quadratic form.
\[
5x^2 + 6x - 8 - 40 = 0
\]
\[
5x^2 + 6x - 48 = 0
\]

#### Step 4: Solve the quadratic equation using the quadratic formula.
The quadratic formula is:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \(a = 5\), \(b = 6\), and \(c = -48\).

\[
x = \frac{-6 \pm \sqrt{6^2 - 4(5)(-48)}}{2(5)}
\]
\[
x = \frac{-6 \pm \sqrt{36 + 960}}{10}
\]
\[
x = \frac{-6 \pm \sqrt{996}}{10}
\]
\[
x = \frac{-6 \pm 2\sqrt{249}}{10}
\]
\[
x = \frac{-3 \pm \sqrt{249}}{5}
\]

Since \(x\) represents a length, it must be positive. Thus:
\[
x = \frac{-3 + \sqrt{249}}{5}
\]

#### Step 7: Calculate the length and width.
\[
\text{Length} = 5x - 4 = 5\left(\frac{-3 + \sqrt{249}}{5}\right) - 4 = -3 + \sqrt{249} - 4 = -7 + \sqrt{249} \, \text{ft}
\]
\[
\text{Width} = x + 2 = \frac{-3 + \sqrt{249}}{5} + 2 = \frac{-3 + \sqrt{249} + 10}{5} = \frac{7 + \sqrt{249}}{5} \, \text{ft}
\]

#### Final Answer for Problem 6:
\[
x = \frac{-3 + \sqrt{249}}{5}, \quad \text{Length} = -7 + \sqrt{249} \, \text{ft}, \quad \text{Width} = \frac{7 + \sqrt{249}}{5} \, \text{ft}
\]

---

Final Answers:

\[
\boxed{
\begin{array}{lll}
1) & x = 2, & \text{Length} = 8 \, \text{ft}, \, \text{Width} = 10 \, \text{ft} \\
2) & x = 6, & \text{Length} = 30 \, \text{yd}, \, \text{Width} = 20 \, \text{yd} \\
3) & x = -2 + \sqrt{70}, & \text{Length} = 4 + \sqrt{70} \, \text{yd}, \, \text{Width} = -4 + \sqrt{70} \, \text{yd} \\
4) & x = 4\sqrt{2}, & \text{Length} = 4\sqrt{2} + 2 \, \text{m}, \, \text{Width} = 4\sqrt{2} - 2 \, \text{m} \\
5) & x = 2\sqrt{34}, & \text{Length} = 2\sqrt{34} + 4 \, \text{in}, \, \text{Width} = 2\sqrt{34} - 4 \, \text{in} \\
6) & x = \frac{-3 + \sqrt{249}}{5}, & \text{Length} = -7 + \sqrt{249} \, \text{ft}, \, \text{Width} = \frac{7 + \sqrt{249}}{5} \, \text{ft} \\
\end{array}
}
\]