Solve and graph the solution set of : 2 x - 9 < 7 and 3 x + 9 ≤ 25 : x ∈ R

Given: 2 x - 9 < 7 and 3 x + 9 ≤ 25 , x ∈ R 2 x < 7 + 9 and 3 x ≤ 25 - 9 ⇒ 2 x < 16 and 3 x ≤ 16 x < 16 2 and x ≤ 16 3 ⇒ x < 8 and x ≤ 5 1 3 ∴ Solution set is x ≤ 5 1 3 , x ∈ R

Solve and graph the solution set of : 2 x - 9 < 7 and 3 x + 9 ≤ 25 : x ∈ R

Given: 2 x - 9 < 7 and 3 x + 9 ≤ 25 , x ∈ R 2 x < 7 + 9 and 3 x ≤ 25 - 9 ⇒ 2 x < 16 and 3 x ≤ 16 x < 16 2 and x ≤ 16 3 ⇒ x < 8 and x ≤ 5 1 3 ∴ Solution set is x ≤ 5 1 3 , x ∈ R

Solve and graph the solution set of 3 x - 2 ≥ 19 or 3 - 2 x ≥ 7 : x ∈ R

3 x - 2 > 19 or 3 - 2 x ≥ - 7 ⇒ 3 x > 19 + 2 or - 2 x ≥ - 7 - 3 ⇒ 3 x > 21 or - 2 x ≥ - 10 ∴ x > 7 or - x ≥ - 5 ∴ x > 7 or x ≤ 5

If 16 a - x a + x 3 = a + x a - x ; show that : a = 3 x .

16 a - x a + x 3 = a + x a - x ⇒ 16 [ ( a - x ) 3 ( a - x ) ] = ( a + x ) ( a - x ) 3 = ( 2 ) 4 ( a - x ) 4 = ( a + x ) 4 we get ⇒ 2 ( a - x ) = a + x 2 a - a = x + 2 x ∴ a = 3 x Hence proved

Solve for x , using the properties of proportionality : 1 + x + x 2 1 - x + x 2 = 62 ( 1 + x ) 63 ( 1 - x ) .

1 + x + x 3 1 - x + x 2 = 62 ( 1 + x ) 63 ( 1 - x ) On cross multiple ⇒ ( 1 + x + x 2 ) ( 1 - x ) ( 1 - x + x 2 ) ( 1 + x ) = 62 63 ⇒ 1 - x + x + x 2 + x 2 - x 3 1 + x - x - x 2 + x 2 + x 3 = 62 63 ⇒ 1 - x 3 1 + x 3 = 62 63 On transposition ⇒ 63 - 63 x 3 = 62 + 62 x 3 ⇒ - 125 x 3 = - 1 x 3 = 1 125 ∴ x = 1 5