Let's work through each equation step-by-step to solve for $x$, and then I will explain the sum and product of the roots.

Equation 6:

$\frac{x - 1}{x + 3} = \frac{2}{x - 3}$ To solve this, we will cross-multiply: $(x - 1)(x - 3) = 2(x + 3)$ Expanding both sides: $(x^2 - 3x - x + 3) = 2x + 6$ Simplifying: $x^2 - 4x + 3 = 2x + 6$ Now, move everything to one side of the equation: $x^2 - 4x - 2x + 3 - 6 = 0$ $x^2 - 6x - 3 = 0$ Now we have a quadratic equation: $x^2 - 6x - 3 = 0$

We can solve this using the quadratic formula: $x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-3)}}{2(1)}$ Simplifying: $x = \frac{6 \pm \sqrt{36 + 12}}{2}$ $x = \frac{6 \pm \sqrt{48}}{2}$ $x = \frac{6 \pm 4\sqrt{3}}{2}$ $x = 3 \pm 2\sqrt{3}$

Thus, the solutions for $x$ are: $x = 3 + 2\sqrt{3}, \quad x = 3 - 2\sqrt{3}$

Sum of the roots: The sum of the roots is $3 + 2\sqrt{3} + 3 - 2\sqrt{3} = 6$.

Product of the roots: The product of the roots is: $(3 + 2\sqrt{3})(3 - 2\sqrt{3}) = 9 - (2\sqrt{3})^2 = 9 - 12 = -3$

Equation 7:

$\frac{x - 4}{x} = \frac{x - 4}{5}$ Here, we can cancel $x - 4$ on both sides (assuming $x \neq 4$): $\frac{1}{x} = \frac{1}{5}$ Cross-multiplying: $x = 5$

The solution for this equation is $x = 5$.

Summary:

• Equation 6:
  • Solutions: $x = 3 + 2\sqrt{3}$ and $x = 3 - 2\sqrt{3}$
  • Sum of the roots: 6
  • Product of the roots: -3
• Equation 7:
  • Solution: $x = 5$

Let me know if you'd like further details or have any questions!

Here are 5 related questions:

1. Can we solve quadratic equations using factoring for equation 6?
2. What would happen if we tried substituting the roots back into the original equations?
3. How can we derive the quadratic formula from a general quadratic equation?
4. Can equation 6 have complex roots in any case?
5. What other methods can be used to solve rational equations like equation 6?

Tip: Always check the domain of the variable when dealing with rational expressions to avoid undefined solutions!