[tex]\frac{3}{2}x + \frac{1}{5} \geq -1[/tex]
[tex](10)\frac{3}{2}x +(10) \frac{1}{5} \geq -1(10)[/tex]
15x + 2 ≥ -10
-2 -2
15x ≥ -12
[tex]\frac{15x}{15} \geq \frac{-12}{15}[/tex]
x ≥ [tex]\frac{-12}{15}[/tex]
x ≥ [tex]\frac{-4}{5}[/tex]
OR
[tex]-\frac{1}{2} x - \frac{7}{3} \geq 5[/tex]
[tex]-(6)\frac{1}{2} x - (6)\frac{7}{3} \geq 5(6)[/tex]
-3x - 14 ≥ 30
+14 +14
-3x ≥ 44
[tex]\frac{-3x}{-3} \leq \frac{44}{-3}[/tex]
[tex]x \leq -\frac{44}{3}[/tex]
Graph: ←-------- [tex]-\frac{44}{3}[/tex] [tex]\frac{-4}{5}[/tex] ----------→
Interval Notation: (-∞, [tex]-\frac{44}{3}][/tex] U [tex][\frac{-4}{5}[/tex], ∞)
Answer:
For given relation the possible values for x can be given by
[tex]x\leq -\frac{44}{3}[/tex] or [tex]x> -\frac{12}{15}[/tex]
Step-by-step explanation:
Given equations are
[tex]\frac{3x}{2}+\frac{1}{5}> -1[/tex] -------(A)
and [tex]-\frac{x}{2}-\frac{7}{3}\geq 5[/tex] -------(B)
First consider equation (A)
[tex]\frac{3x}{2}+\frac{1}{5}> -1 =>\frac{3x}{2}> -\frac{6}{5}[/tex]
=>[tex]x> -\frac{12}{15}[/tex] -------(C)
Now consider equation (B)
[tex]-\frac{x}{2}-\frac{7}{3}\geq 5=>\frac{x}{2}\leq -\frac{22}{3}[/tex]
[tex]x\leq -\frac{44}{3}[/tex] ---------(D)
Using relation (C) and (D)
[tex]x\leq -\frac{44}{3}[/tex] or [tex]x> -\frac{12}{15}[/tex]
Thus for given relation the possible values for x can be given by
[tex]x\leq -\frac{44}{3}[/tex] or [tex]x> -\frac{12}{15}[/tex]
