Solving two step equations using inverse operations | Teaching ... - Free Printable

Great! You've provided a set of equations along with their corresponding flow charts that visually represent the operations applied to the variable $ x $ to reach the final expression. The goal is to use these flow charts to solve the equations, and we’ll go through each one step by step.

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🔧 How Flow Charts Work

- The forward flow (rightward arrows) shows how we build up the expression from $ x $.
- To solve for $ x $, we reverse the process using inverse operations in reverse order (leftward arrows).

Let’s analyze each equation and its flow chart.

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## ✔ 1. Equation: $ 2x + 3 = 8 $

🔹 Flow Chart:

```
x → ×2 → 2x → +3 → 2x+3
↑ ↑ ↑
2.5 5 8
```

🔁 Reverse Process (Solving):

We start from the result 8 and work backwards:

1. Subtract 3:
$ 8 - 3 = 5 $

2. Divide by 2:
$ 5 ÷ 2 = 2.5 $

So, $ x = 2.5 $

✔ Answer: $ x = 2.5 $

> ✔️ Matches the given solution.

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## ✔ 2. Equation: $ 2(x + 3) = 8 $

🔹 Flow Chart:

```
x → +3 → x+3 → ×2 → 2(x+3)
↑ ↑ ↑
1 4 8
```

🔁 Reverse Process (Solving):

Start from 8:

1. Divide by 2:
$ 8 ÷ 2 = 4 $

2. Subtract 3:
$ 4 - 3 = 1 $

So, $ x = 1 $

✔ Answer: $ x = 1 $

> ✔️ Matches the given solution.

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## ✔ 3. Equation: $ x^2 + 3 = 8 $

🔹 Flow Chart:

```
x → square → x² → +3 → x²+3
↑ ↑ ↑
±√5 5 8
```

🔁 Reverse Process (Solving):

Start from 8:

1. Subtract 3:
$ 8 - 3 = 5 $

2. Take square root:
$ \sqrt{5} $ or $ -\sqrt{5} $

So, $ x = \pm\sqrt{5} $

✔ Answer: $ x = \pm\sqrt{5} $

> ✔️ Matches the given solution.

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📌 Summary Table

| Equation | Solution | Steps (Reverse Flow) |
|--------|----------|-----------------------|
| $ 2x + 3 = 8 $ | $ x = 2.5 $ | $ 8 \to -3 \to 5 \to \div 2 \to 2.5 $ |
| $ 2(x + 3) = 8 $ | $ x = 1 $ | $ 8 \to \div 2 \to 4 \to -3 \to 1 $ |
| $ x^2 + 3 = 8 $ | $ x = \pm\sqrt{5} $ | $ 8 \to -3 \to 5 \to \sqrt{} \to \pm\sqrt{5} $ |

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💡 Key Takeaway:

Flow charts help visualize the order of operations. To solve:
- Work backwards using inverse operations.
- Apply them in reverse order (last operation first).
- This method works especially well for linear and simple nonlinear equations.

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Let me know if you'd like to see this method applied to more complex equations!