Question

question
solve for all values of x:
$-\frac{7}{x-1}+4x=-\frac{7x}{x-1}$
answer attempt 1 out of 2
additional solution no solution
$x=$

Explanation:

Step1: Eliminate denominators

Multiply all terms by $x-1$ (where $x
eq 1$):
$$-7 + 4x(x-1) = -7x$$

Step2: Expand the left side

$$-7 + 4x^2 - 4x = -7x$$

Step3: Rearrange to standard quadratic form

Bring all terms to one side:
$$4x^2 - 4x + 7x - 7 = 0$$
$$4x^2 + 3x - 7 = 0$$

Step4: Factor the quadratic

Factor the expression:
$$(4x - 7)(x + 1) = 0$$

Step5: Solve for x

Set each factor equal to 0:
$4x - 7 = 0 \implies x = \frac{7}{4}$
$x + 1 = 0 \implies x = -1$

Step6: Verify solutions

Check $x=-1$ in original equation:
Left side: $-\frac{7}{-1-1} + 4(-1) = \frac{7}{2} - 4 = -\frac{1}{2}$
Right side: $-\frac{7(-1)}{-1-1} = -\frac{7}{2}$
$-\frac{1}{2}
eq -\frac{7}{2}$, so $x=-1$ is extraneous.
Check $x=\frac{7}{4}$ in original equation:
Left side: $-\frac{7}{\frac{7}{4}-1} + 4(\frac{7}{4}) = -\frac{7}{\frac{3}{4}} + 7 = -\frac{28}{3} + 7 = -\frac{7}{3}$
Right side: $-\frac{7(\frac{7}{4})}{\frac{7}{4}-1} = -\frac{\frac{49}{4}}{\frac{3}{4}} = -\frac{49}{12} \times \frac{4}{3} = -\frac{7}{3}$
Both sides are equal, so $x=\frac{7}{4}$ is valid.

Answer:

$x = \frac{7}{4}$