Excluding inequalities

[12345678]

Can somebody explain why you exclude this inequality please…?
P ≥ 39m
A < 99m²
P = 14X + 4

• 14X + 4 ≥ 39
  
• 14X ≥ 35
  
• X ≥ 35/14
  
• X ≥ 5/2

A = 9X² + 6X

• 9X² + 6X < 99
  
• 9X² + 6X – 99 < 0
  
• 3 (3X² + 2X – 33) < 0
  
• 3 (3X + 11)(X – 3) < 0
  
• -11/3 < X < 3

My question: The mark scheme gives the answer as: 5/2 ≤ X < 3
Now I understand where the values come from, but the mark scheme implies that it is obvious to exclude -11/3; anybody know why?

 

[JeffM]

Can somebody explain why you exclude this inequality please…?
P ≥ 39m
A < 99m²
P = 14X + 4

• 14X + 4 ≥ 39
• 14X ≥ 35
• X ≥ 35/14
• X ≥ 5/2

A = 9X² + 6X

• 9X² + 6X < 99
• 9X² + 6X – 99 < 0
• 3 (3X² + 2X – 33) < 0
• 3 (3X + 11)(X – 3) < 0
• -11/3 < X < 3

My question: The mark scheme gives the answer as: 5/2 ≤ X < 3
Now I understand where the values come from, but the mark scheme implies that it is obvious to exclude -11/3; anybody know why?

Your first constraint shows that \(\displaystyle \dfrac{5}{2} \le X.\)

Your second constraint shows that \(\displaystyle - \dfrac{11}{3} < X < 3.\)

This is good work at the mechanics. But what does it MEAN?

Let's consider - 3. Well that certainly satisfies the second constraint, but it does not satisfy the first constraint.

\(\displaystyle - \dfrac{11}{3} < 0 < \dfrac{5}{2}.\)

Consequently, values less than 5/2 have already been excluded by the first constraint, and the lower bound imposed by the second constraint is irrelevant. The upper bound from the second constraint does not contradict the first constraint so it is relevant.

The relevant constraints give you \(\displaystyle \dfrac{5}{2} \le X < 3.\)

Did this help?

 

[12345678]

Your first constraint shows that \(\displaystyle \dfrac{5}{2} \le X.\)

Your second constraint shows that \(\displaystyle - \dfrac{11}{3} < X < 3.\)

This is good work at the mechanics. But what does it MEAN?

Let's consider - 3. Well that certainly satisfies the second constraint, but it does not satisfy the first constraint.

\(\displaystyle - \dfrac{11}{3} < 0 < \dfrac{5}{2}.\)

Consequently, values less than 5/2 have already been excluded by the first constraint, and the lower bound imposed by the second constraint is irrelevant. The upper bound from the second constraint does not contradict the first constraint so it is relevant.

The relevant constraints give you \(\displaystyle \dfrac{5}{2} \le X < 3.\)

Did this help?

Ohh, because the first part says X is greater than or equal to5/2, and the other value is saying it has to be greater than -11/3; so the 2nd bit is reiterating the first part; Cheers!!! and yeah it did help