Solve Linear Inequality: ABS Value(7x-8) <=4x+7

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Discussion Overview

The discussion revolves around solving the linear inequality involving absolute values: ABS value(7x-8) <= 4x+7. Participants explore the steps to solve the inequality, clarify their understanding of the solution set, and address potential errors in reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a method for solving the inequality, leading to the results x >= 5 and x >= 1/11, and questions the correctness of the solution set notation.
  • Another participant identifies a reversal of inequalities in the original solution, suggesting that x <= 5 should be x ≤ 5 instead, and clarifies that the inequality is only reversed when multiplying or dividing by a negative number.
  • A later reply emphasizes the importance of considering cases based on the sign of the expression inside the absolute value, proposing that the first case should yield 8/7 ≤ x ≤ 5 and the second case should yield 1/11 ≤ x ≤ 8/7.
  • Participants discuss the need to take the union of the solution sets rather than combining them directly, indicating that the correct approach involves recognizing the distinct cases for the absolute value.

Areas of Agreement / Disagreement

Participants express differing views on the correct approach to solving the inequality, with some agreeing on the final results while others challenge the methods used to arrive at those results. The discussion remains unresolved regarding the best approach to combine the solution sets.

Contextual Notes

Participants mention the importance of understanding when to reverse inequalities and the implications of the absolute value on the solution process. There are unresolved mathematical steps regarding the correct notation and combination of solution sets.

CanaBra
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Linear inequalities!

I need help solving the following inequality:
ABS value(7x-8) <=4x+7

+/-(7x-8) <=4x+7
7x-8 <=4x+7 -(7x-8) <=4x+7
7x-8-4x <=4x+7-4x -7x+8 <=4x+7
3x-8<=7 -7x+8-4x <=4x+7-4x
3x-8+8<=7+8 -7x-4x+8<=7
3x/3 >= 15 -11x+8<=7-8
x>=5 -11x/-11 >= -1/-11
x>=1/11

Can anyone tell me if this is right?
Also, I don't understand how to write the solution set.
Is this right? 1/11>=x>=5?
Help!
 
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Hmm, my answer has reversed inequalities.

|7x-8| ≤ 4x+7

7x-8 ≤ 4x+7
3x ≤ 15
x ≤ 5

-(7x-8) ≤ 4x+7
-7x+8 ≤ 4x+7
1 ≤ 11x
1/11 ≤ x

So combining the two inequalities:

1/11 ≤ x ≤ 5

EDIT: Your routine to find x >= 1/11 is fine; however, you reversed an inequality in finding x <= 5 when dividing by 3. The inequality will only be reversed if you multiply or divide by a negative number.
 
Last edited:


Thank you very much, I understand now.
 


pbandjay said:
Hmm, my answer has reversed inequalities.

|7x-8| ≤ 4x+7

7x-8 ≤ 4x+7
3x ≤ 15
x ≤ 5

-(7x-8) ≤ 4x+7
-7x+8 ≤ 4x+7
1 ≤ 11x
1/11 ≤ x

So combining the two inequalities:

1/11 ≤ x ≤ 5

EDIT: Your routine to find x >= 1/11 is fine; however, you reversed an inequality in finding x <= 5 when dividing by 3. The inequality will only be reversed if you multiply or divide by a negative number.

Your answer happens to be correct, but I think you need to be a bit more careful with your method here.

Your first case is when 7x-8 is nonnegative which implies x >= 8/7:
|7x-8| ≤ 4x+7

7x-8 ≤ 4x+7
3x ≤ 15
x ≤ 5

That isn't the correct solution for this case. You should have:
8/7 ≤ x ≤ 5

Now your second case is when 7x - 8 is negative, so x ≤ 8/7:
-(7x-8) ≤ 4x+7
-7x+8 ≤ 4x+7
1 ≤ 11x
1/11 ≤ x

Again, that isn't correct. It should be 1/11 ≤ x ≤ 8/7

The union of these two solution sets gives the correct answer. You don't combine your two inequalities. You take the union of the solution sets which, for your sets, would have given the wrong answer.
 


Yes I understand now. Thank you for helping.
 

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