Can we prove that x>1 implies y<1 if 3x+2y≤5 for real numbers x and y?

[bean29]

Suppose x and y are real numbers and 3x+2y≤5. Prove that if x>1 then y<1

 

[statdad]

What would happen if $x > 1$ and $y \ge 1$ were both true?

 

[bean29]

Not real sure what would happen.

 

[mfb]

This is not an acceptable way to ask for help at homework problems. In addition, it is in the wrong forum.

 

[blue_raver22]

To prove that if x>1 then y<1, we can use a proof by contradiction.

Assume that x>1 and y≥1.

Since 3x+2y≤5, we can substitute y≥1 into the inequality to get:

3x+2(1)≤5

Simplifying, we get:

3x+2≤5

Subtracting 2 from both sides, we get:

3x≤3

Dividing both sides by 3, we get:

x≤1

However, this contradicts our initial assumption that x>1. Therefore, our assumption must be false and y<1.

Hence, if x>1 then y<1.