Taking Absolute Cases 2 (1 Viewer)

Lukybear

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This i have a problem with. I was taught to multiply by (x+y)^2 when the x is in the denominator. So for this question of absolutes



The solutions for this, did not require multiplying denomnator by (x+5)^2, but rather took 2 cases. I was wondering what the rules for this is?
 

Carrotsticks

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I believe that since the denominator is always positive, you can bring it to the other side without worrying about a change of sign.

Once you have brought it to the other side, you can apply the '2 cases' principle associated with absolute values.
 

Drongoski

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The final inequality you solve as you do in 2U maths.
 

cutemouse

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or you could square both sides, since both sides of the inequality are positive.
 

Lukybear

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Thxs for all kind answers. Especially vafa, for all hard work.
 

Lukybear

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What about absolute Qs with variables on both sides?
I.e.

|x-5|<5x+9
 

gurmies

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|x - 5| < 5x + 9

Solve for |x - 5| = 5x + 9

x - 5 = 5x + 9 or/ x - 5 = -5x - 9

x = -7/2 or/ x = -2/3

After testing x > -2/3, -7/2 < x < -2/3 and x < -7/2; you can deduce that x > -2/3
 

Trebla

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What about absolute Qs with variables on both sides?
I.e.

|x-5|<5x+9
| x - 5 | < 5x + 9
=> - (5x + 9) < x - 5 < 5x + 9

For this particular question, there is the condition that 5x + 9 ≥ 0 since
5x + 9 > | x - 5 | ≥ 0
hence x ≥ - 9/5 is a restriction

Splitting the inequalities into two parts
x - 5 > - 5x - 9
6x > - 4
x > - 2/3
OR
x - 5 < 5x + 9
4x > - 14
x > - 7/2

To satisfy the three inequalities simultaneously would give x > - 2/3
 

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