Let's solve the system of equations using the substitution method:

We are given:

1. $x^2 + y^2 = 26$
2. $x - y = 4$

Step 1: Solve for one variable in terms of the other

From the second equation $x - y = 4$, we can solve for $x$:

$x = y + 4$

Step 2: Substitute this expression for $x$ into the first equation

Now, substitute $x = y + 4$ into the first equation $x^2 + y^2 = 26$:

$(y + 4)^2 + y^2 = 26$

Step 3: Expand and simplify the equation

First, expand $(y + 4)^2$:

$(y + 4)^2 = y^2 + 8y + 16$

So, the equation becomes:

$y^2 + 8y + 16 + y^2 = 26$

Combine like terms:

$2y^2 + 8y + 16 = 26$

Now, subtract 26 from both sides:

$2y^2 + 8y - 10 = 0$

Step 4: Simplify the quadratic equation

We can divide the entire equation by 2 to simplify:

$y^2 + 4y - 5 = 0$

Step 5: Solve the quadratic equation

We will now factor the quadratic equation:

$y^2 + 4y - 5 = (y + 5)(y - 1) = 0$

This gives two possible solutions for $y$:

$y + 5 = 0 \quad \Rightarrow \quad y = -5$ $y - 1 = 0 \quad \Rightarrow \quad y = 1$

Step 6: Solve for $x$

Now, substitute these $y$-values into the expression $x = y + 4$:

• When $y = -5$, $x = -5 + 4 = -1$
• When $y = 1$, $x = 1 + 4 = 5$

Step 7: Check the solutions

We now check both pairs $(x, y)$:

• For $x = -1$ and $y = -5$:

$x^2 + y^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26 \quad \text{(True)}$

• For $x = 5$ and $y = 1$:

$x^2 + y^2 = 5^2 + 1^2 = 25 + 1 = 26 \quad \text{(True)}$

Both pairs satisfy the original system of equations.

Final Solutions:

The solutions are: $(x, y) = (-1, -5) \quad \text{and} \quad (x, y) = (5, 1)$

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Would you like more details or explanations about any part of the solution?

Here are 5 related questions to explore:

1. How do we factor quadratic equations in general?
2. What are other methods to solve systems of equations (substitution, elimination, etc.)?
3. How can we verify if a solution to a system of equations is correct?
4. What is the geometric interpretation of the system $x^2 + y^2 = 26$ and $x - y = 4$?
5. How do we solve a quadratic equation when it does not factor easily?

Tip: Always check your solutions by substituting them back into the original equations to ensure accuracy!