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.
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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?
 

Thread 'Significant figures for inherently bound values'

I have been insisting to my statistics students that for probabilities, the rule is the number of significant figures is the number of digits past the leading zeros or leading nines. For example to give 4 significant figures for a probability: 0.000001234 and 0.99999991234 are the correct number of decimal places. That way the complementary probability can also be given to the same significant figures ( 0.999998766 and 0.00000008766 respectively). More generally if you have a value that...
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