MHB How to Select the Correct X-Value for Testing Inequalities?

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To solve the inequality 2x - 7 < 11, the correct value of x must be less than 9. Testing with x = 0 confirms this, as it satisfies the inequality. However, simply checking one value is insufficient to prove that all numbers less than 9 work; it only demonstrates that at least one does. The discussion emphasizes the importance of selecting appropriate x-values for testing inequalities to ensure comprehensive validation. Ultimately, the conclusion is that x must be less than 9 for the original inequality to hold true.
mathdad
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Solve the inequality.

2x - 7 < 11

2x < 11 + 7

2x < 18

x < 18/2

x < 9

Correct?
 
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It's easy to check. If x is any number less than 9 then 2x is less than 18 so that 2x- 7 is less than 11. Yes, that is correct.
 
Good to be correct.

2x - 7 < 11

The value of x must be less than 9 to make the original inequality a true statement.

Let x = 0

2(0) - 7 < 11

0 - 7 < 11

-7 < 11

This is true. So, x < 9 is correct.
 
RTCNTC said:
Good to be correct.

2x - 7 < 11

The value of x must be less than 9 to make the original inequality a true statement.

Let x = 0

2(0) - 7 < 11

0 - 7 < 11

-7 < 11

This is true. So, x < 9 is correct.
No, that is not an appropriate way to check. That shows that there exist a number, less than 9, that satisfies the equation. It does not show that every number less than 9 satisfies it.

For example, suppose you had arrived at the incorrect conclusion that the solution was x< 5. Taking x= 0, which is still less than 5, would arrive at the same result.
 
Not every number less than 9 can be used to show that the original inequality is true. How does one select the correct x-value for testing?
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...

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