Question

what is the solution set of the equation 2/x - 3x/x+3 = x/x+3

Answer 1

Answer:

x = -1; x = ³/₂

Step-by-step explanation:

 2/x - 3x/(x + 3) = x/(x+3)     Subtract x/(x+ 3) from each side

 2/x - 4x/(x+ 3)  = 0             Divide each side by 2

  1/x - 2x/(x + 3) = 0             Multiply each side by x(x+ 3)

        x + 3 - 2x² = 0             Multiply each side by -1

        2x² - x - 3 = 0              Factor the quadratic

    (2x - 3)(x + 1) = 0             Find the zeroes

2x - 3 = 0       x + 1 = 0

    2x = 3             x = -1

      x = ³/₂

The solution set is x = -1, x = ³/₂.

The graph shows the x-intercepts at (-1, 0) and (1.5,0) and the vertical asymptotes at x = -3 and x = 0.

Answer 2

Solving a quadratic equation, and paying attention to the domain, the solution set of the equation is:

$S = \left\{-1, \frac{3}{2}\right\}$

The equation given is:

$\frac{2}{x} - \frac{3x}{x + 3} = \frac{x}{x + 3}$

Placing everything on the same side of the equality:

$\frac{2}{x} - \frac{3x}{x + 3} - \frac{x}{x + 3} = 0$

$\frac{2}{x} - \frac{4x}{x + 3} = 0$

Applying the least common factor:

$\frac{2(x + 3) - 4x^2}{x(x + 3)} = 0$

Then:

$\frac{-4x^2 + 2x + 6}{x(x + 3)} = 0$

In a fraction, the denominator cannot be zero, thus $x = 0$ and $x = 3$ cannot belong to the solution set.

The solutions are the values of x for which:

$-4x^2 + 2x + 6 = 0$

Simplifying by -2:

$2x^2 - x - 3 = 0$

Which is a quadratic equation with coefficients $a = 2, b = -1, c = -3$.

Then, applying Bhaskara:

$\Delta = (-1)^2 - 4(2)(-3) = 25$

$x_{1} = \frac{1 + \sqrt{25}}{2(2)} = \frac{3}{2}$

$x_{2} = \frac{1 - \sqrt{25}}{2(2)} = -1$  

Thus, the solution set is:

$S = \left\{-1, \frac{3}{2}\right\}$

A similar problem is given at https://brainly.com/question/13136492