Let's solve each equation step by step.

1. Solving the first equation:

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

First, let's eliminate the denominators by multiplying the entire equation by $X(X+4)$:

$X(X+4) \left(\frac{3}{X} + \frac{5}{X+4}\right) = 1 \cdot X(X+4)$

Simplifying:

$3(X + 4) + 5X = X(X + 4)$

Expanding both sides:

$3X + 12 + 5X = X^2 + 4X$

Combine like terms:

$8X + 12 = X^2 + 4X$

Bring all terms to one side of the equation:

$X^2 + 4X - 8X - 12 = 0$

Simplify:

$X^2 - 4X - 12 = 0$

Now, solve the quadratic equation using the quadratic formula:

$X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -4$, and $c = -12$:

$X = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-12)}}{2(1)}$

Simplifying:

$X = \frac{4 \pm \sqrt{16 + 48}}{2} = \frac{4 \pm \sqrt{64}}{2} = \frac{4 \pm 8}{2}$

So:

$X_1 = \frac{12}{2} = 6, \quad X_2 = \frac{-4}{2} = -2$

2. Solving the second equation:

$\frac{1}{X} + \frac{1}{X + 3} = \frac{1}{2}$

Multiply the entire equation by $2X(X + 3)$ to eliminate the denominators:

$2X(X + 3) \left(\frac{1}{X} + \frac{1}{X + 3}\right) = 2X(X + 3) \cdot \frac{1}{2}$

Simplifying:

$2(X + 3) + 2X = X(X + 3)$

Expanding:

$2X + 6 + 2X = X^2 + 3X$

Combine like terms:

$4X + 6 = X^2 + 3X$

Bring all terms to one side:

$X^2 + 3X - 4X - 6 = 0$

Simplify:

$X^2 - X - 6 = 0$

Solve the quadratic equation:

$X = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -1$, and $c = -6$:

$X = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)}$

Simplifying:

$X = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm \sqrt{25}}{2} = \frac{1 \pm 5}{2}$

So:

$X_1 = \frac{6}{2} = 3, \quad X_2 = \frac{-4}{2} = -2$

Summary of Solutions:

• From the first equation: $X = 6$ and $X = -2$
• From the second equation: $X = 3$ and $X = -2$

Note: Since $X = -2$ appears in both equations, it's important to check that it doesn't lead to a denominator of zero in either equation (which it does not), so all solutions are valid.

Would you like more details on any step, or do you have further questions?

Related Questions:

1. What happens if a solution causes a denominator to become zero?
2. How do you derive the quadratic formula from a standard quadratic equation?
3. How can you check the solutions of a quadratic equation?
4. What are the properties of the discriminant in the quadratic formula?
5. How do you graph the solutions of a quadratic equation?

Tip:

When solving rational equations, always check for extraneous solutions by substituting back into the original equation.