Answer:
[tex]x < -3[/tex]
Interval notation: (-∞, 3)
Step-by-step explanation:
Given inequality:
[tex]-4 \left(\dfrac{5}{2}+\dfrac{3}{2}x\right) > 8[/tex]
Divide both sides by -4 (remembering to change the direction of the inequality sign, since we are dividing by a negative):
[tex]\implies \dfrac{ -4 \left(\dfrac{5}{2}+\dfrac{3}{2}x\right)}{-4} > \dfrac{8}{-4}[/tex]
[tex]\implies \dfrac{5}{2}+\dfrac{3}{2}x < -2[/tex]
Subtract ⁵/₂ from both sides:
[tex]\implies \dfrac{5}{2}+\dfrac{3}{2}x-\dfrac{5}{2} < -2-\dfrac{5}{2}[/tex]
[tex]\implies \dfrac{3}{2}x < -\dfrac{9}{2}[/tex]
Multiply both sides by 2:
[tex]\implies \dfrac{2 \cdot 3}{2}x < -\dfrac{2 \cdot 9}{2}[/tex]
[tex]\implies \dfrac{\diagup\!\!\!\!2 \cdot 3}{\diagup\!\!\!\!2}x < -\dfrac{\diagup\!\!\!\!2 \cdot 9}{\diagup\!\!\!\!2}[/tex]
[tex]\implies 3x < -9[/tex]
Divide both sides by 3:
[tex]\implies \dfrac{3x}{3} < \dfrac{-9}{3}[/tex]
[tex]\implies x < -3[/tex]
Learn more about inequalities here:
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