Question

Which of the following is the solution to the compound inequality below? 3/2x+1/5≥-1 or -1/2 x-7/3≥ 5

Answer 1

$\frac{3}{2}x + \frac{1}{5} \geq -1$

$(10)\frac{3}{2}x +(10) \frac{1}{5} \geq -1(10)$  

15x + 2 ≥ -10

       -2     -2  

     15x ≥ -12

      $\frac{15x}{15} \geq \frac{-12}{15}$

         x ≥ $\frac{-12}{15}$

         x ≥ $\frac{-4}{5}$

OR

$-\frac{1}{2} x - \frac{7}{3} \geq 5$

$-(6)\frac{1}{2} x - (6)\frac{7}{3} \geq 5(6)$

-3x - 14 ≥ 30

     +14    +14

     -3x ≥ 44

       $\frac{-3x}{-3} \leq \frac{44}{-3}$

       $x \leq -\frac{44}{3}$        

Graph:  ←--------  $-\frac{44}{3}$      $\frac{-4}{5}$ ----------→

Interval Notation: (-∞, $-\frac{44}{3}]$ U $[\frac{-4}{5}$, ∞)

Answer 2

Answer:

For given relation the possible values for x can be given by

$x\leq -\frac{44}{3}$     or   $x> -\frac{12}{15}$

Step-by-step explanation:

Given equations are

$\frac{3x}{2}+\frac{1}{5}> -1$ -------(A)

and $-\frac{x}{2}-\frac{7}{3}\geq 5$ -------(B)

First consider equation (A)

$\frac{3x}{2}+\frac{1}{5}> -1 =>\frac{3x}{2}> -\frac{6}{5}$

=>$x> -\frac{12}{15}$     -------(C)

Now consider equation (B)

$-\frac{x}{2}-\frac{7}{3}\geq 5=>\frac{x}{2}\leq -\frac{22}{3}$

$x\leq -\frac{44}{3}$    ---------(D)

Using relation (C) and (D)

$x\leq -\frac{44}{3}$     or   $x> -\frac{12}{15}$

Thus for given relation the possible values for x can be given by

$x\leq -\frac{44}{3}$     or   $x> -\frac{12}{15}$