# Question: Is 2x less than y+1?

## Comment on Is 2x less than y+1?

### In stmnt 1, even 2x can be y

In stmnt 1, even 2x can be y<2x<y+1 like logic given for stmnt 2. Please explain

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 condition y<y+1 is valid only if Y is positive right ?

### 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

I still dont understand why y + 2 in statement 2 can go on the either side. If the question is asking if 2x < y + 1, and you know 2x < y + 2 to be true, how can it be that y + 1 go either side?

### 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

When to rephrase the Target Question . Like Here the target question can be rephrased as 2x-y<1, And that led me to the wrong conclusion. Please answer?

### 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

### Hey Guys,

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

Perhaps it was just an intuition that without the +1, the question would be Is 2x < y which would imply is x < y/2. However, adding 1 to y implies that the question is now asking if x <= y/2.
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

Can you please verify my approach to this DS question?
(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

### Your solution is perfect! I

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

Cheers,
Brent

### Hi Brent,

Hi Brent,

For Inequality and Algebra DS questions, I've noticed that sometimes we rephrase the target question and sometimes we rephrase the statements. I'm getting confused on when to rephrase the target question vs the statement. Is there a logic or strategy behind when we look to rephrase the target question vs the indivivual statements?

### Whenever you have a somewhat

Whenever you have a somewhat complicated algebraic expression, equation, or any quality (regardless of whether it's part of a target question or a statement), it's always going to be useful to simplify (i.e., rephrase) that information.

### Hi,

Hi,

I rephrased the question to be is X<(Y+1)/2 by dividing by 2 on both side. Then I rewrote as, is X<(Y/2)+(1/2)?

Statement 1 is sufficient. I added 1/2 to both side and as X < X+1/2 so X < Y/2 + 1/2

Statement 2 is not as X-(1/2) < (Y/2) + (1/2) can't tell us if X < (Y/2) + (1/2) ==> But this wasn't easy to tell...

I always try to rephrase. Thanks,

### That's a great approach!

That's a great approach!
The thing I like most about it is that the rephrased target question looks a lot like each of the two statements.

### Hi Brent, getting a bit

Hi Brent, getting a bit confused with cross multipliction in inequality. Without knowing x is +- in both statements here, how could we cross multipy it? Have I missed something here? Thanks Brent

### You're absolutely right to

You're absolutely right to say that, unless we know the sign of a VARIABLE, we can't multiply both sides of an inequality by that variable.
However, for this problem I'm not multiplying both sides by any variables.
In fact, I'm not even cross multiplying.

Statement 1 says: x < y/2
To simplify this inequality, I'm going to multiply both sides by 2.
2 is NOT a VARIABLE.
Also, since we know that 2 is definitely a positive number, we can safely multiply both sides by 2.
When we do this we get: 2x < y

Does that help?

### Ah I see what you mean. In

Ah I see what you mean. In other words, if we have a positive number in either side of an inequation then it's safe to cross multiply? So long it's not both unknow variables? Thanks Brent

### Yes, if you're certain both

Yes, if you're certain both values are positive, you can cross multiply (when it comes to inequalities)

### Great thanks Brent for

Great thanks Brent for confirmation.