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Sagot :
Answer:
{ x ∣ x > 1 } or x ∈ ( 1 , ∞ )
Step-by-step explanation:
2 x + 3 > 7
⇒ 2 x > 4
(subtract 3 from both sides)
2 x + 3 > 7
⇒ x > 2
(divide both sides by 2)
2 x + 9 > 11
⇒ 2 x > 2
(subtract 9 from both sides)
2 x + 9 > 11
⇒ x > 1
(divide both sides by 2)
if an x -value is greater than 2, it will automatically be greater than 1. Thus, the solution set for 2 x+ 3 > 7 is a subset of the one for 2 x + 9 >11 .That means, all we need to do here is list the solution set for 2 x + 9 > 11 , and we're done.
The solution set we need is simply "all x such that x is greater than 1", or { x ∣ x > 1 }
or
x ∈ ( 1 , ∞ )
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