Question

Help me solve this problem please?

Answer 1

`(x+7)/(x-8)\ge3`

The first thing we have to see is that this equation is indeterminate when the value of its denominator is equal to zero. We call this value critical value

`\begin{gathered} x-8=0 \\ x=8\to\text{critical value} \end{gathered}`

Now we will solve the inequality

`\begin{gathered} (x+7)/(x-8)\ge3 \\ x+7\ge3(x-8) \\ x+7\ge3x-3(8) \\ 7+24\ge3x-x \\ 31\ge2x \\ x\le(31)/(2) \end{gathered}`

Using the value of the inequality and the critical value found earlier, we can determine the range of this

[tex]\begin{gathered} 8