The image you've uploaded contains two equations that appear to be related to transforming an equation into a quadratic form. I'll extract the mathematical content and help you solve the problem:

Equations on the board:

1. $\frac{2}{X} + \frac{5}{X+4} = 1$
2. $\frac{3}{X} + \frac{1}{X-3} = \frac{1}{2}$

Goal:

Transform these equations into quadratic equations to solve for the values of $X$.

Let's start with solving the first equation step-by-step and move on to the second equation afterward.

Equation 1: $\frac{2}{X} + \frac{5}{X+4} = 1$

I'll solve this first by clearing the denominators and forming a quadratic equation.

from sympy import symbols, Eq, solve

Define the symbol

X = symbols('X')

First equation: (2/X) + (5/(X+4)) = 1

equation_1 = Eq(2/X + 5/(X + 4), 1)

Solve for X

solutions_1 = solve(equation_1, X) solutions_1The solutions to the first equation $\frac{2}{X} + \frac{5}{X+4} = 1$ are:

$X = \frac{3}{2} - \frac{\sqrt{41}}{2} \quad \text{or} \quad X = \frac{3}{2} + \frac{\sqrt{41}}{2}$

Now, let's proceed to solve the second equation:

Equation 2: $\frac{3}{X} + \frac{1}{X-3} = \frac{1}{2}$

I'll work through it next.# Second equation: (3/X) + (1/(X-3)) = 1/2 equation_2 = Eq(3/X + 1/(X - 3), 1/2)

Solve for X

solutions_2 = solve(equation_2, X) solutions_2The solutions to the second equation $\frac{3}{X} + \frac{1}{X-3} = \frac{1}{2}$ are:

$X = 2 \quad \text{or} \quad X = 9$

If you have any questions or need further details on how these solutions were derived, feel free to ask!

Related Questions:

1. How do we handle quadratic equations when they result in complex solutions?
2. What is the general process for clearing denominators in rational equations?
3. How can we verify the solutions by substituting them back into the original equations?
4. Can these types of equations be solved graphically as well?
5. How does the discriminant of a quadratic equation help in predicting the nature of the solutions?

Tip:

Always check for extraneous solutions when solving rational equations by substituting back into the original equation.