Detailed walkthrough of solving a nonlinear system of equations using algebraic substitution to find the points (5, -1) and (1, -5).

Let's solve the system of equations algebraically using substitution, as shown in your image, and carefully walk through each step to verify the solution.

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🔷 Given System:

1. $ x^2 + y^2 = 26 $  → (Equation 1: a circle)
2. $ x - y = 6 $    → (Equation 2: a line)

We are told to use substitution.

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🔹 Step 1: Solve one equation for one variable

From Equation 2:
$$
x - y = 6
\Rightarrow y = x - 6
$$

Now substitute this expression for $ y $ into Equation 1.

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🔹 Step 2: Substitute into the first equation

Replace $ y $ with $ x - 6 $ in $ x^2 + y^2 = 26 $:

$$
x^2 + (x - 6)^2 = 26
$$

Now expand $ (x - 6)^2 $:

$$
(x - 6)^2 = x^2 - 12x + 36
$$

So:

$$
x^2 + (x^2 - 12x + 36) = 26
$$

Simplify:

$$
x^2 + x^2 - 12x + 36 = 26 \\
2x^2 - 12x + 36 = 26
$$

Subtract 26 from both sides:

$$
2x^2 - 12x + 10 = 0
$$

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🔹 Step 3: Simplify and factor

Divide entire equation by 2:

$$
x^2 - 6x + 5 = 0
$$

Now factor:

$$
(x - 5)(x - 1) = 0
$$

So:

$$
x = 5 \quad \text{or} \quad x = 1
$$

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🔹 Step 4: Find corresponding $ y $ values

Recall: $ y = x - 6 $

- If $ x = 5 $:
$ y = 5 - 6 = -1 $ → Point: $ (5, -1) $

- If $ x = 1 $:
$ y = 1 - 6 = -5 $ → Point: $ (1, -5) $

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🔹 Step 5: Check solutions in original equations

#### Check $ (5, -1) $:

1. $ x^2 + y^2 = 5^2 + (-1)^2 = 25 + 1 = 26 $ ✔
2. $ x - y = 5 - (-1) = 5 + 1 = 6 $ ✔

#### Check $ (1, -5) $:

1. $ x^2 + y^2 = 1^2 + (-5)^2 = 1 + 25 = 26 $ ✔
2. $ x - y = 1 - (-5) = 1 + 5 = 6 $ ✔

Both points satisfy both equations.

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✔ Final Answer:

$$
\boxed{(5, -1) \quad \text{and} \quad (1, -5)}
$$

This matches the answer shown in your image.

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📝 Summary:

We used substitution:
- Solved the linear equation for $ y $
- Substituted into the quadratic equation
- Solved the resulting quadratic
- Found two solutions
- Verified both in the original system

✔ The system has two solutions: $ (5, -1) $ and $ (1, -5) $.