Question

$0 = \frac{15x^2 - 8x}{x-1} - 3x+1$

Answer 1

The solutions are $x = \frac{1}{2}$ and $x = -\frac{1}{6}$.

Answer 2

The given equation is $0 = \frac{15x^2 - 8x}{x-1} - 3x+1$.

First, we need to identify the domain of the variable $x$. The denominator cannot be zero, so $x-1 \neq 0$, which means $x \neq 1$.

Rearrange the equation to combine the terms: $\frac{15x^2 - 8x}{x-1} = 3x-1$

Multiply both sides by $(x-1)$, assuming $x \neq 1$: $15x^2 - 8x = (3x-1)(x-1)$

Expand the right side: $(3x-1)(x-1) = 3x(x-1) - 1(x-1) = 3x^2 - 3x - x + 1 = 3x^2 - 4x + 1$

Substitute this back into the equation: $15x^2 - 8x = 3x^2 - 4x + 1$

Move all terms to one side to form a quadratic equation: $15x^2 - 3x^2 - 8x + 4x - 1 = 0$ $12x^2 - 4x - 1 = 0$

This is a quadratic equation in the form $ax^2 + bx + c = 0$, with $a=12$, $b=-4$, and $c=-1$. We can solve this using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substitute the values of $a$, $b$, and $c$: $x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(12)(-1)}}{2(12)}$ $x = \frac{4 \pm \sqrt{16 + 48}}{24}$ $x = \frac{4 \pm \sqrt{64}}{24}$ $x = \frac{4 \pm 8}{24}$

This gives two possible solutions for $x$: $x_1 = \frac{4 + 8}{24} = \frac{12}{24} = \frac{1}{2}$ $x_2 = \frac{4 - 8}{24} = \frac{-4}{24} = -\frac{1}{6}$

Finally, we must check if these solutions satisfy the domain restriction $x \neq 1$. For $x_1 = \frac{1}{2}$, $\frac{1}{2} \neq 1$. So, $x = \frac{1}{2}$ is a valid solution. For $x_2 = -\frac{1}{6}$, $-\frac{1}{6} \neq 1$. So, $x = -\frac{1}{6}$ is a valid solution.

Both solutions are valid.