To solve the given system of equations using the substitution method, follow these steps:

The system of equations is:

1. $xy = 5$
2. $x^2 + y^2 = 26$

Step 1: Solve for $y$ in terms of $x$ from the first equation

From $xy = 5$, we isolate $y$:

$y = \frac{5}{x}$

Step 2: Substitute $y = \frac{5}{x}$ into the second equation

Substitute into $x^2 + y^2 = 26$:

$x^2 + \left( \frac{5}{x} \right)^2 = 26$

Simplify $\left( \frac{5}{x} \right)^2$:

$x^2 + \frac{25}{x^2} = 26$

Step 3: Eliminate the fraction by multiplying through by $x^2$

To clear the denominator, multiply the entire equation by $x^2$ (assuming $x \neq 0$):

$x^2 \cdot x^2 + x^2 \cdot \frac{25}{x^2} = 26x^2$

Simplify:

$x^4 + 25 = 26x^2$

Step 4: Rearrange into standard form

Bring all terms to one side to form a quadratic equation in terms of $x^2$:

$x^4 - 26x^2 + 25 = 0$

Let $z = x^2$ (substitution to simplify). Then the equation becomes:

$z^2 - 26z + 25 = 0$

Step 5: Solve the quadratic equation

Solve for $z$ using the quadratic formula $z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -26$, and $c = 25$:

$z = \frac{-(-26) \pm \sqrt{(-26)^2 - 4(1)(25)}}{2(1)}$

Simplify:

$z = \frac{26 \pm \sqrt{676 - 100}}{2}$

$z = \frac{26 \pm \sqrt{576}}{2}$

$z = \frac{26 \pm 24}{2}$

This gives two solutions:

1. $z = \frac{26 + 24}{2} = 25$
2. $z = \frac{26 - 24}{2} = 1$

Step 6: Back-substitute $z = x^2$

Recall $z = x^2$. Therefore:

1. $x^2 = 25$ → $x = \pm 5$
2. $x^2 = 1$ → $x = \pm 1$

Step 7: Solve for $y$ corresponding to each $x$

Using $y = \frac{5}{x}$, find $y$ for each value of $x$:

1. For $x = 5$:
   $y = \frac{5}{5} = 1$ Solution: $(x, y) = (5, 1)$

2. For $x = -5$:
   $y = \frac{5}{-5} = -1$ Solution: $(x, y) = (-5, -1)$

3. For $x = 1$:
   $y = \frac{5}{1} = 5$ Solution: $(x, y) = (1, 5)$

4. For $x = -1$:
   $y = \frac{5}{-1} = -5$ Solution: $(x, y) = (-1, -5)$

Final Solutions:

The solutions to the system are:

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

Would you like a graph or further verification for these solutions?

Related Questions:

1. How do you solve systems of non-linear equations graphically?
2. Can you solve the system using elimination instead of substitution?
3. How do you verify solutions to a system of equations?
4. What are real-world applications of solving such systems?
5. How does the substitution method differ from other algebraic techniques?

Tip:

Always simplify and rearrange equations carefully to avoid calculation errors, especially when dealing with non-linear systems.