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

[CanaBra]

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!

 

[pbandjay]

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.

 

[CanaBra]

Thank you very much, I understand now.

 

[LCKurtz]

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.

 

[pbandjay]

Yes I understand now. Thank you for helping.