If 0

Question

If  0<2x+3y<10  and  -10<3x+2y<0 , then which of the following must be true?

I.   x<0  
II.   y<0 
III.   x<y

A. I only 
B. II only 
C. I & II 
D. I & III 
E. I, II, &III

chetan2u · Answer

MathRevolution , pl correct the OA..
 0<2x+3y<10  and  -10<3x+2y<0 ............ 3x+2y<0<2x+3y..............(2x+2y)+x<0<(2x+2y)+y

from above you can make out that when x is added to 2x+2y it becomes negative and when to same value 2x+2y, y is added, it becomes positive..
so x<0 and y>0. Also x<y
I and III are correct ..
D..

Just looking at CHOICES you can eliminate all but 2..

0<2x+3y<10  and  -10<3x+2y<0 
If both x and y are positive, 3x+2y can never be NEGATIVE..... so  -10<3x+2y<0  is not possible
If both x and y are negative, 2x+3y can never be POSITIVE..... so  0<2x+3y<10  is not possible
therefore both x and y are of opposite sign..
1) BOTH I and II cannot be true.... eliminate C and E
2) if y<0, x cannot be <y otherwise both x and y become <0..

so either I and III are true or ONLY II is true
in choices only B and D are left

niks18 · Answer

0<2x+3y<10 -----------(1) -10<3x+2y<0 ----------(2), Multiply this inequality by -1 =>0<-3x-2y<10 , add this inequality with inequality (1), to get 0 y-x>0 or y>x . Hence Statement III must be true Also from inequality (2) we know that 3x+2y<0 . This implies both x & y are of opposite signs. As we have established that III must be true so y cannot be negative. Hence x<0 . Therefore statement I must be true Answer: option D

ScottTargetTestPrep · Answer

We see that x and y can’t be both positive or both negative. If x and y are both positive, then the second inequality will not hold. Similarly, if x and y are both negative, then the first inequality will not hold. Therefore, we must consider two separate cases: (1) x is negative and y is positive and (2) x is positive and y is negative.

Case 1. Let’s assume that x is negative and y is positive.

For example, If x = -5 and y = 5, we see that we do have 0 < 2x + 3y < 10 and -10 < 3x + 2y < 0.

Case 2. Now let’s assume that x is positive and y is negative. We see that the absolute value of y must be greater than the absolute value of x in order for the second inequality to hold.

For example, if x = 2, y has to be less than -3 in order to have -10 < 3x + 2y < 0. However, in that case, the first inequality will never hold since 2x + 3y will be negative. Thus we can’t have x as positive and y as negative.

Thus it must be true that x is negative and y is positive and in that case we also have x < y. Thus, Statements I and III must be true.

Answer: D

MathRevolution · Answer

=>

Label the inequalities as follows:
0<2x+3y<10 --- (1)
-10<3x+2y<0 --- (2)
We consider each statement individually.

Statement I:
Multiplying (1) by -2 yields -20 < -4x – 6y < 0, and multiplying (2) by 3 yields -30 < 9x + 6y < 0. Adding these inequalities gives -50 < 5x < 0 or -10 < x < 0.
This statement is true.

Statement II:
Multiplying (1) by 3 yields 0 < 6x + 9y < 30, and multiplying (2) by -3 yields 0 < -6x – 4y < 20. Adding these inequalities gives 0 < 5y < 50 or 0 < y < 10. 
This statement is false.

Statement III:
Multiplying (1) by – 1 yields -10 < -2x – 3y < 0. Adding this to inequality (2) yields -20 < x – y < 0.
This implies that x < y, and statement III is true.

Therefore, the answer is D.
Answer: D

SergejK · Answer

Actually, if we just look at the equations that are given, we see that if y becomes smaller and x becomes larger, the before positive solution becomes negative. So we can infer that x must be negative and y must be positive, as both inequalities are each additions. hence, we can rule out S2 and logically, it follows that y>x.

repellatquas · Answer

Alternatively,
Looking at the equations,
3x+2y < 0 & 2x+3y > 0
We can deduce that -> 3x + 2y < 2x+3y
Therefore, x < y (this has to be true)
Thus, option III is always true. Basis that, we can eliminate, A,B,C as they do not include option III

We now only that to check for option II since option I is included in both D and E choices.
We can do that simply by finding one value for which the option would not hold true.

Since II says y< 0, we can check for simpler values like -1.
If y=-1, and x = -2 (as x has to be less than y), then it should hold true that 2x+3y>0
On checking, we see that 2(-2) + 3(-1) = -1, which is not greater than 0. This negates the eq given in question.
Eliminate E

Answer - D.

egmat · Answer

This problem is about combining inequalities strategically to isolate x, y, and (x - y).

We're given:
  Inequality 1:   0  <  2 x +  3 y <  10 
  Inequality 2:   -10  <  3 x +  2 y <  0

Checking Statement I (x < 0): 
Multiply Inequality  1  by  2 :  0  <  4 x +  6 y <  20 
Multiply Inequality  2  by  3 :  -30  <  9 x +  6 y <  0

To subtract, flip the first:  -20  < -( 4 x +  6 y) <  0 
Now add it to the second:  -30  + ( -20 ) <  5 x <  0  +  0 
Result:  -50  <  5 x <  0 , so  -10  < x <  0 .
x is definitely negative.  Statement I is TRUE.

Checking Statement II (y < 0): 
Multiply Inequality  1  by  3 :  0  <  6 x +  9 y <  30 
Multiply Inequality  2  by  2 :  -20  <  6 x +  4 y <  0

Flip the second:  0  < -( 6 x +  4 y) <  20 
Add to the first:  0  +  0  <  5 y <  30  +  20 
Result:  0  <  5 y <  50 , so  0  < y <  10 .
y is actually positive!  Statement II is FALSE. This eliminates choices B, C, and E.

Checking Statement III (x < y): 
We already know x <  0  and y >  0 . A negative number is always less than a positive number, so x < y automatically.  Statement III is TRUE.

Statements I and III must be true.

Answer: D

Key principle:  When you have two compound inequalities with the same variables, you can multiply them by constants and add/subtract to isolate individual variables — just like solving a  system of equations , but remembering to flip the inequality when multiplying by a negative.

If 0<2x+3y<10 and -10<3x+2y<0, then which of the following must be tru

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