Sure, let's solve the given equation step-by-step.
We are given the equation:
[tex]\[
\frac{x + 2}{y} = \frac{y}{x - 3}
\][/tex]
To solve this equation, let's first eliminate the fractions by cross-multiplying:
[tex]\[
(x + 2)(x - 3) = y \cdot y
\][/tex]
Simplify the left-hand side:
[tex]\[
(x + 2)(x - 3) = x^2 - 3x + 2x - 6 = x^2 - x - 6
\][/tex]
So, our new equation is:
[tex]\[
x^2 - x - 6 = y^2
\][/tex]
This is a quadratic equation in terms of [tex]\(x\)[/tex]. Let's solve for [tex]\(x\)[/tex] in terms of [tex]\(y\)[/tex]. We rewrite the equation as:
[tex]\[
x^2 - x - 6 - y^2 = 0
\][/tex]
This is a standard quadratic equation of the form [tex]\(ax^2 + bx + c = 0\)[/tex] where:
[tex]\[
a = 1, \quad b = -1, \quad c = -6 - y^2
\][/tex]
We solve this quadratic equation using the quadratic formula:
[tex]\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\][/tex]
Substitute [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
[tex]\[
x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6 - y^2)}}{2 \cdot 1}
\][/tex]
[tex]\[
x = \frac{1 \pm \sqrt{1 + 24 + 4y^2}}{2}
\][/tex]
[tex]\[
x = \frac{1 \pm \sqrt{25 + 4y^2}}{2}
\][/tex]
Therefore, we have two solutions for [tex]\(x\)[/tex]:
[tex]\[
x = \frac{1 + \sqrt{25 + 4y^2}}{2}
\][/tex]
[tex]\[
x = \frac{1 - \sqrt{25 + 4y^2}}{2}
\][/tex]
Thus, the solutions to the original equation are:
[tex]\[
\left(\frac{1 + \sqrt{25 + 4y^2}}{2}, y\right)
\][/tex]
[tex]\[
\left(\frac{1 - \sqrt{25 + 4y^2}}{2}, y\right)
\][/tex]
Hence, the solutions are:
[tex]\[
\left(\frac{1 - \sqrt{4y^2 + 25}}{2}, y\right) \quad \text{and} \quad \left(\frac{\sqrt{4y^2 + 25} + 1}{2}, y\right)
\][/tex]
So, the final solution set for the given equation is:
[tex]\[
\left(\frac{1 - \sqrt{4y^2 + 25}}{2}, y\right) \quad \text{and} \quad \left(\frac{1 + \sqrt{4y^2 + 25}}{2}, y\right)
\][/tex]
