GMAT Math

in the book that i'm using, i feel that they dont explain how to solve absolute value equations very well, so i'd like confirmation from an instructor to make sure i'm doing it correctly and that i have covered all scenarios

22 - |y+14| = 20

i first solve to get the inequality as the only thing on one side of the equation, then i know that there are 2 options

-|y+14| = 20 - 22
-|y+14| = -2
|y+14| = 2

1.

y+14 = 2
y = -12

2.a.

y + 14 = -2
y = -16

2.b. since the value inside the absolute value brackets can be negative, i mutliply everything within the absolute value brackets by negative 1

-y - 14 = 2
-y = 16
y = 16

i know that 2.a. and 2.b. actually are equal to each other, but is one way the correct way of doing absolute value equations?

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separately for inequalities, i did some googling and came across a site that said that if the greater than/less than sign is facing one way you have to do one thing and if it's the other way you can solve it by flipping the sign.

like i think for less than or equal to it said to solve it like this:

ax - b <= c, then -c <= ax - b <= c

and for greater than or equal to, it said for one solution you would solve it normally, then you find the 2nd solution by changing the direction of the inequality sign

rather than having to remember when to flip signs, i'm trying to think of a hard and fast rule i can apply to both absolute value equations and inequalities. again, could an instructor verify that what i'm doing is correct?


|2x - 5| <= 7

1.

2x - 5 <= 7
2x <= 12
x <= 6

2.a.

2x - 5 >= -7
2x >= -2
x >= -1

2.b.

-2x + 5 <=7
-2x <= 2
x >= -1

again, i know that 2.a. and 2.b. are equivalent, but is it correct that if i make the 7 negative, i then reverse the way the inequality is pointing? or is the correct way to do it the 2.b. way, or vice versa. i'm just very confused about inequalities

then lastly,

|-3x + 2 | > 7

1.

-3x + 2 >7
-3x > 5
3x < 5
x < (5/3)


2.a.

-3x + 2 < -7
-3x < -9
3x > 9
x > 3

2.b.

3x - 2 > 7
3x > 9
x > 3


i know that step 1 is always needed for the 3 examples i gave. i think i prefer to do step 2.b. for all equations because there is no flippng of signs, and it's consistent that you have to always make everything in the absolute value bracket negative. just need confirmation.


thank you very much!

Anurag@Gurome wrote:

dhlee922 wrote:2.a.

y + 14 = -2
y = -16

2.b. since the value inside the absolute value brackets can be negative, i mutliply everything within the absolute value brackets by negative 1

-y - 14 = 2
...

i know that 2.a. and 2.b. actually are equal to each other, but is one way the correct way of doing absolute value equations?

That's what you already did in 2.a.
2.a ---> (y + 14) = -2 <---- Multiply both sides by -1 --> -y - 14 = 2 <--- 2.b

2.b is completely redundant as it is a different version of 2.a

By definition of |x|,

• |x| = x if x ≥ 0 and |x| = -x if x < 0

Hence, here

• if (y + 14) ≥ 0, |y + 14| = (y + 14) = 2 ---> y = -12
  if (y + 14) < 0, |y + 14| = -(y + 14) = 2 ---> y = -16

I hope you understand the possible two different scenarios by logic not by merely remembering the process.

The same logic applies to all the other examples also.

thanks anurag for the replies. i'm obviously rusty in simple math skills, but i thought you always have to multiply both sides of the equation, but the 2.a. example only multiplies one side of the equation. is that what you always do for inequality questions?

i initially meant that when solving these i would not do both 2.a. and 2.b. I just found that 2a and 2b were 2 ways to think of it and i would choose one of them, and i believe your 2nd reply is similarly stating that 2b is the way to go based on your 2nd reply

i really appreciate you explaining the logic behind it as that has cleared things up.

|-3x + 2 | > 7

1.

-3x + 2 >7
-3x > 5
3x < 5
x < (5/3)





For this example why is x < (5/3) as opposed to x <(-5/3)


-3x+2 > 7

-3x> 5

(divide by negative 3)

x< -5/3


where did i go wrong? thanks!

Brent@GMATPrepNow wrote:

rkav wrote:|-3x + 2 | > 7

1.

-3x + 2 >7
-3x > 5
3x < 5
x < (5/3)

For this example why is x < (5/3) as opposed to x <(-5/3)
-3x+2 > 7
-3x> 5
(divide by negative 3)
x< -5/3
where did i go wrong? thanks!

If |x| > k (where k is greater than 0) then x > k or x , -k

So, if |-3x + 2 | > 7, then -3x + 2 > 7 or -3x + 2 < -7
Tackle each one separately:
-3x + 2 > 7
-3x > 5
x < -5/3

-3x + 2 < -7
-3x < -9
x > 3

So, x < -5/3 or x > 3

Cheers,
Brent

Thank you Brent.

So just to be clear one of the answers is x< negative 5/3?

Thanks Brent. That is what I got as well.


Thank you for clearing this up for me.

rkav, i meant the parenthese to equal a negative sign, i think that's where your confusion is coming from

hi, i have a follow up question regarding absolute values

when is |x-4| equal to 4-x?

so i start with 2 equations, first one is:

x-4 = 4-x
2x = 8
x = 4

then 2nd equation i have:

-(x-4) = 4-x
-x+4 = 4-x
these 2 equations are equal, so i thought you would have only one solution, but the book says the answer is x <= 4

how come i cant apply step 2.b. from the initial question at the top to this problem?

Your methods work well when you start with an equation or inequality. But here you are just asked a question about some expression. There is no equation given here, so they expect you to fall back on the definition of absolute value notation.

|(stuff)| means (stuff), when stuff is positive, and -(stuff) when stuff is negative, for example |4| is (4), but |-4| is -(-4).

In your example, |x-4| will equal 4-x whenever "stuff" is negative, ie in this case when (x-4) is negative. That happens when x<4.

The "zero" solution is interesting to think about too. I would keep my eye pealed for answers playing on the edge of any constraints I derived. You can use the answers to tip you off and then test the extreme values.

Cheers,

misterholmes