Where Did I Go Wrong When Solving Inequalities With Negative Numbers?

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SUMMARY

The discussion focuses on solving the inequality 4 > 19 - 3x and highlights a common mistake when dividing by a negative number. The correct steps involve recognizing that dividing both sides by -3 requires reversing the inequality sign, resulting in the correct solution of x > 5. Participants emphasize the importance of checking each step with test values to confirm the validity of the inequality.

PREREQUISITES
  • Understanding of basic algebraic inequalities
  • Knowledge of the properties of inequalities when multiplying or dividing by negative numbers
  • Ability to perform basic arithmetic operations
  • Familiarity with testing values to verify inequality solutions
NEXT STEPS
  • Study the rules of inequalities, particularly the effects of multiplying or dividing by negative numbers
  • Practice solving various types of inequalities, including linear and quadratic forms
  • Learn how to use test values effectively to validate solutions to inequalities
  • Explore advanced topics in algebra, such as systems of inequalities and their graphical representations
USEFUL FOR

Students learning algebra, educators teaching inequality concepts, and anyone looking to strengthen their understanding of solving inequalities with negative numbers.

paulb203
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Homework Statement
Solve for x
4>19-3x
Relevant Equations
N/A
My attempt:

4>19-3x

Subtract 19 from both sides:

-15 > -3x

Divide both sides by -3:

5 > x

Switch sides (change sign):

x < 5

! But Maths Genie tells me the answer is x>5

Where have I gone wrong?
 
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You did not divide by -3 correctly.
 
paulb203 said:
Homework Statement: Solve for x
4>19-3x
Relevant Equations: N/A

My attempt:

4>19-3x

Subtract 19 from both sides:

-15 > -3x

Divide both sides by -3:

5 > x
This is wrong. If you multiply or divide by a negative number, then you have to reverse the '>' sign. Thus it has to be 5 < x.

paulb203 said:
Switch sides (change sign):

x < 5
You switched sign AND sides, so you didn't change anything. 5 < 6 and 6 > 5 are the same.
paulb203 said:
! But Maths Genie tells me the answer is x>5

Where have I gone wrong?
 
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paulb203 said:
Homework Statement: Solve for x
4>19-3x
Relevant Equations: N/A

My attempt:

4>19-3x

Subtract 19 from both sides:

-15 > -3x

Divide both sides by -3:

5 > x

Switch sides (change sign):

x < 5

! But Maths Genie tells me the answer is x>5

Where have I gone wrong?
One way to find your error yourself is to check each step with numbers that should meet the inequality.

We start with ##4 > 19-3x##.
Try ##x = 4##. We have ##19 -3x = 7##, so the inequality fails.
Try ##x = 6##. We have ##19 -3x = 1##, so the inequality holds.

Your next inequality was ##-15 > -3x##.
Try ##x = 4##. We have ##-3x = -12##, so the inequality fails.
Try ##x = 6##. We have ##-3x = -18##, so the inequality holds. You are still all right.

Then you have ##5 > x##. By trying ##x = 4## and ##x = 6##, we can see that at this step things have gone wrong, and it should be ##5 < x##.
 
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Hill said:
You did not divide by -3 correctly.
Thanks, Hill

Are you talking about the issue with the sign?

Is my division correct on its own, putting the issue with the sign to one side for the moment;

-15 / -3 = 5 ?

-3x / -3 = x ?
 
paulb203 said:
Are you talking about the issue with the sign?
Yes.
 
“If you multiply or divide by a negative number, then you have to reverse the '>' sign. “

Ah, thanks, I think I remember this now.

Is this correct:

Example:

-15 > -18

both sides by -3

5 > 6 (incorrect, because they are now positive numbers, and 5 is obv. not greater than 6)
 
Thanks, peroK
 
paulb203 said:
“If you multiply or divide by a negative number, then you have to reverse the '>' sign. “

Ah, thanks, I think I remember this now.

Is this correct:

Example:

-15 > -18

both sides by -3

5 > 6 (incorrect, because they are now positive numbers, and 5 is obv. not greater than 6)
Yes, exactly. The acid test is that ##x > 0## is equivalent to ##-x < 0## (multiplying or dividing both sides by ##-1##).

You have to be careful algebraically as well. The following only applies when ##a > 0##:
$$ab > ac \ \Rightarrow \ b > c \ \ \ (a > 0)$$
 

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