# How do solve x-10/(x-1)>=4 algebraically?

##### 1 Answer
Mar 27, 2017

-1 ≤ x < 1 uu x ≥ 6

#### Explanation:

Combine the left hand side into one fraction: $\frac{x \left(x - 1\right)}{x - 1} - \frac{10}{x - 1} = \frac{{x}^{2} - x - 10}{x - 1}$. The equation then becomes (x^2-x-10)/(x-1)-4≥0. We combine the fraction again to (x^2-5x-6)/(x-1)≥0

Now, there are two possible ways to solve it. The first way is to multiply both sides by $x - 1$, dividing the problem into two cases depending on the sign of $x - 1$, solve a quadratic equation, and draw a sign diagram to determine the solution.

The simpler method would be to directly draw the sign diagram. We first find the zeros of the numerator and denominator. For the numerator, the zeroes are $6$ and $- 1$. Since the zeros are not repeated, the equation changes signs at each zero. The zeros divide the equation into different ranges of where the signs can be. By quickly substituting values in, we determine that when $x < - 1 \cup x > 6$, the equation is positive. The equation is negative when $- 1 < x < 6$.

For the denominator, when $x > 1$, it is positive. When $x < 1$, it is negative.

From this, we can determine four ranges where for its sign:

• between $- \infty$ and $- 1$ (the numerator is positive, the denominator is negative; the fraction is negative)
• between $- 1$ and $1$ (the numerator is negative; the denominator is negative; the fraction is positive)
• between $1$ and $6$ (numerator is negative; the denominator is positive; the fraction is negative)
• between $6$ and $\infty$ (numerator is positive; the denominator is positive; the fraction is positive)

We find that the fraction is greater than $0$ when it is positive, or when $- 1 < x < 1 \cup x > - 6$. It is equal to zero when the numerator is $0$, or when $x = - 6 \cup x = - 1$.

The final solution is thus -1 ≤ x < 1 uu x ≥ 6.