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## Comment on

Is 2x less than y+1?## In stmnt 1, even 2x can be y

## In your suggested inequality

In your suggested inequality (y<2x<y+1), y is less than 2x. This is not true.

In statement 1, we first concluded that 2x < y (after we multiplied both sides by 2). So y is definitely greater than 2x.

Since y < y+1, our only conclusion can be 2x < y < y + 1

## the condition y<y+1 is valid

## The inequality y<y+1 is valid

The inequality y<y+1 is valid for ALL y-values.

When we add 1 to any value, that value becomes greater.

If you`re not convinced, try testing some y-values.

Try y = 0

We get: 0 < 0 + 1

Simplify: 0 < 1 (works)

Try y = 5

We get: 5 < 5 + 1

Simplify: 5 < 6 (works)

Try y = -3

We get: -3 < -3 + 1

Simplify: -3 < -2 (works)

And so on.

## I still dont understand why y

## All we know for certain is

All we know for certain is that y+1 < y+2

Statement two tells us (indirectly) that 2x < y+2

So where does y+1 fit here? We know that it must go to the left of y+2, but where?

Let's test some values...

CASE A: x = 2.75 and y = 10

This satisfies statement 2 (2x < y+2)

With these x- and y-values, y+1 fits BETWEEN 2x and y+2

We get: 2x < y+1 < y+2

So, we can conclude that 2x < y+1

CASE B: x = 2.75 and y = 4

This satisfies statement 2 (2x < y+2)

With these x- and y-values, y+1 fits to the LEFT of 2x

We get: y+1 < 2x < y+2

So, we can conclude that y+1 < 2x

Since the two CASES provide different answers to the target question, statement 2 is not sufficient.

## When to rephrase the Target

## There's nothing wrong with

There's nothing wrong with rephrasing the target question as "Is 2x-y < 1?" (as you have done)

For statement 1, we get: 2x < y

Subtract y from both sides to get: 2x - y < 0

This means 2x-y is NEGATIVE, which means 2x-y is definitely less than 1 (using your rephrased target question)

So, statement 1 is sufficient

For statement 2, we get: 2x < y + 2

Subtract y from both sides to get: 2x - y < 2

So, it COULD be the case that 2x - y = 1.5, in which case 2x-y > 1 (using your rephrased target question)

Or it COULD be the case that 2x - y = 0, in which case 2x-y < 1 (using your rephrased target question)

Since we cannot answer the rephrased target question with certainty, statement 2 is NOT sufficient

Answer: A

## Hey Guys,

When I rephrased, I got "Is x <= Y/2 and statement 1 addressed that easily and statement 2 was ambiguous.

## I like the fact that you're

I like the fact that you're looking for ways to rephrase the target question, but I'm not sure how you rephrased "Is 2x < y + 1?" to get "Is x ≤ y/2?"

Can you explain?

Cheers,

Brent

## Perhaps it was just an

Did I confuse you?...lol

## I was a bit confused :-)

I was a bit confused :-)

Your rephrased target question worked THIS TIME, but I'd probably avoid such approximations in the future; they could lead to the wrong answer.

## Can you please verify my

(or suggest any alternative?)

Is x < 5?

1) x^2 > 5

2) x^2 + x <5

my approach:

1)x > 5^(1/2) and -5^(1/2),

x = 6 and -6, so insufficient

2) x < x^2 + x <5,

so, sufficient

Thanks in advance

## Your solution is perfect! I

Your solution is perfect! I can't think of a faster/better approach.

Nice work.

Cheers,

Brent

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