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## Comment on

1/10 and 1/2## We can also solve for 1) as

we can rewrite expression in 1) as

10>4x+2>2, this means, 4x+2 can be 3,4,5,6,7,8 or 9

if we solve above, we get

8>4x>0

divide by 4

2>x>0,

that leave us option: x=1.

Can we do following?

1/x<1/y<1/z then x>y>z, i think so.

## Fantastic!

Fantastic!

Your suggested rule (if 1/x <1/y < 1/z then x > y > z) is true AS LONG AS x, y, and z are all positive or all negative.

For example, 1/(-2) < 1/(-3) < 1/2, however we can't say that -2 > -3 > 2

## Thank you so much.

I really liked your videos. The way that you solves and explains little facts are really amazing as these little facts can trick in simple questions.

Cheers

Atul

## Great video. Question: on

## Once we know that 4x+2 equals

Once we know that 4x+2 equals one of the integers from 3 to 9, we can automatically eliminate all of the ODD integers, since 4x will be EVEN and when we add 2, the result is EVEN.

This leaves us with 4, 6 or 8 top test.

## Thank you! 3-9 I think you

## Brent,

Please help me, I solved the first stated as below.

1/10 < 1/4x+2 < 1/2

0.1 < 1/4x+2 < 0.5

Since there is no integer that lies between 0.1 & 0.5, I declared this statement to be insufficient.

## Be careful. We are told that

Be careful. We are told that x is an integer. We are not told that 1/(4x+2) is an integer.

For example, ix x = 1, then 1/(4x+2) = 1/6 = 0.166...

## Brent,

Exercise 224 (Problem Solving - OG 2017)

Could you explain me in a different way from OG 2017 answer solution?

Thank you in advance.

Cheers,

Pedro

## At the risk of appearing to

At the risk of appearing to "pass the buck," I'd like to direct you to a very rich discussion of this question here: http://www.beatthegmat.com/og-13-ps-218-is-this-prob-really-doable-withi...

In particular, Mitch (GMATGuruNY) provides a concise solution.

Also, Ceilidh (from Manhattan Prep) explains why this might be a great candidate for guessing and moving on.

If you have any questions about those various solutions, I'm happy to respond. I just don't think I'd do much better than Mitch's solution.

## Hey Brent, on statement 2,

## You only need check for

You only need check for extraneous roots when dealing with equations with square roots and equations with absolute value.

Cheers,

Brent

## To avoid testing all the

## That works too. Nice job!

That works too. Nice job!

## Hi Brent, for statement 1,

Such that 1/10 < 1/4x+2 < 1/2 becomes:

1) 1/10 < 1/4x+2 and 2) 1/4x+2 < 1/2

This gives us 1) x < 2 and 2) x > 0

If we combine 1) and 2) again to bring back us back to to the original equality, we get:

0<x<2. Since x is an integer, x must be 1.

Just want to know if this approach works for all inequality questions with 2 inequality signs.

Thanks.

## That's a perfectly valid

That's a perfectly valid solution. Nice work.

Cheers,

Brent

## In the video question above,

x could be 0 - couldn't it?

## x is, indeed, an integer.

x is, indeed, an integer.

However, x = 0 is not a solution to either statement.

Let's see why:

Statement 1) 1/10 < 1/(4x + 2) < 1/2

Plug in x = 0 to get: 1/10 < 1/(4(0) + 2) < 1/2

Simplify to get: 1/10 < 1/2 < 1/2

Since it is NOT the case that 1/2 < 1/2, x = 0 is not a possible value of x.

Statement 2) x² + 5x - 6 = 0

Plug in x = 0 to get: 0² + 5(0) - 6 = 0

Simplify to get: 0 + 0 - 6 = 0 (doesn't work)

So, 0 is not a possible value of x.

Does that help?

Cheers,

Brent

## Hi. I tried to actually solve

## Let's see how that approach

Let's see how that approach will work.

Given: 1/10 < 1/(4x+2) < 1/2

Multiply all sides by (4x+2) to get : (4x+2)/10 < 1 < (4x+2)/2

Multiply all sides by 10 to get : (4x+2) < 10 < 5(4x+2)

Expand to get : 4x + 2 < 10 < 20x + 10

We can treat this a two separate inequalities:

4x + 2 < 10 and 10 < 20x + 10

Let's solve each one.

Take: 4x + 2 < 10

Subtract 2 from both sides to get: 4x < 8

Divide both sides by 4 to get: x < 2

Take: 10 < 20x + 10

Subtract 10 from both sides to get: 0 < 20x

Divide both sides by 20 to get: 0 < x

If we combine x < 2 and 0 < x, we get: 0 < x < 2

Since x is an INTEGER, x must equal 1

Does that help?

Cheers,

Brent

## Hi Brent! Could you tell me

In respect of statement 1 can we assume 2 situations i.e. when x is positive and when x is negative.

When x is positive the equation becomes 10<4x+2<2 which is not possible and hence can be discarded.

When x is negative the equation becomes 10>4x+2>2 which simplified gives us 2>x>0 and the only integer value possible in this case is x=1.

Is this a valid approach?

## That's a good idea, but that

That's a good idea, but that's not quite how the inequality works.

You write: "When x is positive the equation becomes 10 < 4x+2 < 2 which is not possible and hence can be discarded"

If x is positive, then it must be the case that 2 < 4x+2 < 10 (not 10 < 4x+2 < 2)

Since, x is an integer, we can be certain that x = 1.

So, we can't discard that scenario.

---------------------------------

Conversely, the statement "When x is negative the equation becomes 10>4x+2>2 which simplified gives us 2>x>0 and the only integer value possible in this case is x=1" is contradictory.

If x is negative, then we can't then conclude that x = 1 (a positive number)

Does that help?

Cheers,

Brent

## Hi Brent. Please help me to

1/10 < 1/(4x+2) < 1/2

I really struggled with x being in the denominator so I tried the following:

10 > 4x +2 > 2.

Then follows, 8 > 4x > 0 , and finally 2 > x > 0. As the only possible integer between these values is 1. x equals 1.

Is that correct or should I stop doing such things?

Cheers.

Alex

## That approach is perfectly

That approach is perfectly valid.

In fact, it's pretty much the same as my approach :-)

You concluded that 10 > 4x +2 > 2

I concluded that 4x + 2 = 3, 4, 5, 6, 7, 8, or 9 (since x must be an integer)

Cheers,

Brent

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