Chapter 5.3, Problem 25PPS

To determine

To solve the inequality:

  $8x-27\le -x+18$

And check the solution.

Answer to Problem 25PPS

  $x\le 5$

Explanation of Solution

Given:

  $8x-27\le -x+18$

Concept Used:

• To check the answer solution for the given inequality

1. First check the solution at end point by converting it into an equation.
2. If the solution is $x<m$, then pick any value of x less than m and check whether it satisfies the inequality.
3. If the solution is $x>m$, then pick any value of x greater than m and check whether it satisfies the inequality.

Properties of inequality:

1. $b\le c\Rightarrow ab\le ac,\text{ if }a\ge 0$
2. $b\le c\Rightarrow ab\ge ac,\text{ if }a\le 0$

Calculation:

Let the number be x, eight times of a number x is 8x and subtract it by 27 can be written as:

  $8x-27$

Now, then 8x − 27 value is no more than the negative of that number is written as:

  $\begin{array}{l}8x-27\le -x+18\\ \Rightarrow 8x-27-\left(-x+18\right)\le 0\\ \Rightarrow 9x-45\le 0\\ \Rightarrow 9\left(x-5\right)\le 0\end{array}$

Divide each side by 9 to solve the value of x:

  $\begin{array}{c}\frac{9\left(x-5\right)}{9}\le \frac{0}{9}\\ \left(x-5\right)\le 0\\ x\le 5\end{array}$

Thus, the inequality is true for x is less than and equals to 5, thus the solution of given inequality in interval form is:

  $\left(-\infty ,5\right]$

Now, to check the answer solution for the given inequality first check the solution at end point by converting it into an equation. Putting x equals to 5 in left hand side and right hand side:

  $\begin{array}{l}\text{At }x=5:\\ 8\left(5\right)-27\le -\left(5\right)+18\\ \Rightarrow 40-27\le 13\\ \Rightarrow 13\le 13\end{array}$

Put any value any value of x less than 5, say 4 and check whether it satisfies the inequality, as

  $\begin{array}{c}x=4:8x-27\le -x+18\\ 8\left(4\right)-27\le -\left(4\right)+18\\ 32-27\le -\left(4\right)+18\\ 5\le 14\end{array}$

Since, the condition satisfied thus solution of inequality is correct.