2x+3 is greater than or equal to 7 OR 2x+9 is greater than 11. Solve for x.

{ 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 , ∞ )

y + 11 = (1/4)(x + 16)

step-by-step explanation: