Question

How do you solve and write the following in interval notation: $\left| 1+5x \right|\le 11$?

Answer 1

In this type of question when there is a modulus given there are two cases that can be formed one with positive equation and the other with negative and by solving those two equation we will get two inequalities with which we can write the interval of x.

Complete step-by-step solution:
In the above type of question we will be removing the modulus that is present and then we will form two different inequalities one with positive value and the other one with negative value i.e.

& 1+5x\le 11.....\left( 1 \right) \\\ & 1+5x\ge -11.....\left( 2 \right) \\\ \end{aligned}$$ So we will solve each inequality individually i.e. let us first solve equation 1 in which we will subtract 1 on both sides of inequality and then after subtracting we will divide it by 5 on both sides of inequality from solving this we will be getting the upper limit of the interval i.e. x cannot be greater then the solution we got from solving equation 1. $$\begin{aligned} & \Rightarrow 1+5x-1\le 11-1 \\\ & \Rightarrow 5x\le 10 \\\ & \Rightarrow \dfrac{5x}{5}\le \dfrac{10}{5} \\\ & \Rightarrow x\le 2 \\\ \end{aligned}$$ So after solving the first equation we finally get the value $$x\le 2$$ which states that the max value of x cannot exceed 2 which gives us our upper limit of the interval. Now to find the lower limit of the interval we will use equation 2. In equation two again we will subtract 1 on both sides of inequality then we will divide both sides of inequality by 5 to get the final answer as lower limit of inequality. $$\begin{aligned} & \Rightarrow 1+5x-1\ge -11-1 \\\ & \Rightarrow 5x\ge -12 \\\ & \Rightarrow \dfrac{5x}{5}\ge \dfrac{-12}{5} \\\ & \Rightarrow x\ge -2.4 \\\ \end{aligned}$$ In this we for the lower limit of interval as -2.4 i.e. value of x cannot be less than -2.4 which gives us our interval as $$x\in \left[ -2.4,2 \right]$$ **The interval of the given limit is $$x\in \left[ -2.4,2 \right]$$** **Note:** In this type remember to use the modulus function i.e. opening the modulus results in two different functions which upon solving will result in our final answer i.e. the interval of the function that has been asked in the question.