ax + by=10, ay + bx=12. which one is greater here, x or y ?

ax + by=10
bx + ay=12

subtracting these 2

(b-a)x + (a-b)y =2
or
(a-b)x - (a-b)y = -2
(a-b)(x-y) = -2

statement 1:
a>b and
(a-b)(x-y) = -2
(+)*(-)= -ve
therefore x-y<0
hence x<y
sufficient

statement 2:
a/b>1
(a-b)/b > 0
now a-b can be positive or negative
hence insufficient

hence A

After replying to a PM, I thought I have to make a few comments on this question.

This question is about comparing two numbers \(x\) and \(y\).
The best way to do this, is to look at the difference of the two:
if \(x - y = 0\), the two numbers are equal
if \(x - y > 0\), then \(x\) is greater than \(y\)
if \(x - y < 0\), then \(x\) is less than \(y\)

The "trap" here is the given system of the two equations. We don't need to solve it.
The two numbers are totally irrelevant, the only important thing is that 10 < 12.
Therefore, we can write that
\(ax + by < ay + bx\), or \(ax - ay - bx + by < 0\).

After factorisation we get
\(a(x - y) - b(x - y) = (x - y)(a - b) < 0\).

This means that \(x - y\) and \(a - b\) have opposite signs and that neither one can be 0.

From (1), \(a > b\) or \(a - b > 0\), so, necessarily \(x - y < 0\) and \(y\) is greater than \(x\).
Sufficient.

From (2), if \(b > 0\), then \(a > b\) (multiply the given inequality by \(b > 0\), direction preserved).
Then \(a - b > 0\), so \(x - y\) must be negative and \(x\) is smaller than \(y\).
If \(b < 0\), then \(a < b\) (multiply the given inequality by \(b < 0\), direction is reversed).
Then \(a - b < 0\), so \(x - y\) must be positive and \(x\) is greater than \(y\).
Not sufficient.

Hence, Answer A.

Remark:
When dealing with the inequality \(a/b > 1\), since \(b\) cannot be 0, we can multiply by \(b^2\), which is for sure positive and get
\(ab > b^2\)
This leads to \(ab - b^2\) > 0 or \(b(a - b) > 0\).
Now, we can easily see that if \(b > 0\), then \(a > b\).
When \(b < 0\), then \(a < b\).

Suggestion:
An inequality involving a fraction with unknown sign of the denominator can always be multiplied by the square of that
denominator, and the direction of the inequality is preserved.
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algebraic solution..very good.. i knew it. but i love geometry that"s why went that way....
you did a fantastic job here too... i appreciate it... so nice

Can you please explain a little bit more about the statement 2. How did you arrive at (a-b) ? I solved it as follows:
\(\frac{a}{b} >1\)
Multiply both sides by b.
\(a>b\)
This is the same as Statement 1. Hence the answer is D. Please clarify the answer.

highlited portion is wrong a/b>1
as you are multiplying both sides with b
there will be 2 cases
case 1: b>0
if b is +ve then equality sign doesnt change and hence you can come to a>b
case 2: b<0
if b is -ve then equality sign changes reslting equation will become a<b

now since we are not aware whether b is positive or negative so it is advisable not to multiply with b rather bring all thing on one side
therefore \(\frac{a}{b}-1>0\)
solving this \(\frac{(a-b)}{b}>0\)
now if b is-ve then a-b needs to be -ve hence a-b<0==> a<b
if b is +ve then a-b needs to be +ve hence a-b>0==> a>b

hope this helps

Bunnuel can you attempt this one. Not entirely convinced of the solution. I may be missing something.

Solving for \(x\)and \(y\)in terms of \(a\) and \(b\) and then making\(x>y\) results in the following inequality

\((a+b)^2(a-b) < 0\)

Now \(a > b\) is NOT sufficient for this inequality because we can have \(5 > -5\) making \((a+b)^2(a-b) = 0\)
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blueseas your reply is more than satisfactory...really fine

Actually you are missing out on an important point. You have ASSUMED that x>y and also, the fact IS that a>b. Thus, in your final expression which is :

\((a+b)^2(a-b) < 0\), all the terms are non negative, and it can never be negative. Thus, the assumption that x>y is wrong.

Another method: From F.S 1, we have

ax + by=10
ay + bx=12

Subtract both, and we get : \((a-b)x+y(b-a) = -2 \to (a-b)(x-y) =-2\). Now, given that a>b, i.e. a-b>0. Thus, \((x-y)<0 \to x<y.\) Sufficient.

From F.S 2, if you notice closely, the only difference between this and the previous fact statement is that in this case, we don't know if a>b or not. We are just given that \(\frac{a}{b}>1\). Depending upon the sign of b, we could end up with a>b OR a<b.

Thus, using the same equation which we got above, i.e. (a-b)(x-y) =-2, for a>b we will end up with x>y(just as above) and for a<b, we will have (a-b)<0. Thus, x-y will have to be positive in this case. Thus, Insufficient.

Hope this helps.
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\(ax + by = 10\)
\(bx + ay = 12\)
Solving for\(x\) and \(y\)we have

\(x = (12b-10a)/(b^2-a^2)\\
y = (10b- 12a)/(b^2-a^2)\)

Since the question stem says \(x > y\) we do the same and try to find the relation between \(a\) and \(b\).

\((12b-10a)/(b^2-a^2) > (10b- 12a)/(b^2-a^2)\)

We get \((b+a)/[(b+a)(b-a)] > 0\)------------(I)

From I we cannot cancel \((b+a)\)from numerator and denominator unless we know\((b+ a) =! 0\), since the question doesnot mention that \((b + a)\) is not equal to \(0\) we cannot conclude that for \(x > y\) we need \(b > a\) or vice versa. For example if we have \(b = 5 & a = -5\) here we have\(b > a\)but the expression I is not defined.

Hence believe A is NOT the answer.
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hi abhi ,

great effort for solving for X and Y.

As the final equation is (as per your calculation)= \((b+a)/(b+a)(b-a) > 0\)
b+a cant be equal to zero since statement 1 says that a>b so it means they are not equal hence a+b cant be equal to zero.
for a+b=0...the above equation is not defined as you will have 0 in denominator.

moreover i dont understand why you are saying that you cannot cancel a+b....there is no problem in cancelling out a+b.
hence the end equation becomes:
\((1)/(b-a) > 0\)
now it depends on which one is greater ...and statement one clearly says a>b hence
so you are getting definite ANSWER NO.

i hope i have not puzzled you more.

Bluseas lets say \(a = 5\) \(b = -5\) here we have \(a > b\) but \(a + b = 0\).

Can you cancel a +b from numerator and denominator ?
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my bad..

mistake from my side.

but i will say when you are deriving X and Y ..in terms of a and b ...DENOMINATOR cannot be zero ...if that is zero then in that case X and Y are not defined ...and since GMAT deals with real numbers so we have to understand that a+b or a-b cant be zero.

now i would like to know what was your concern...as you were saying that you cannot cancel a+b.
can you just elaborate.

thanks

Add the given equations :

\(ax + by = 10\)
\(bx + ay = 12\)

You get x(a+b)+y(a+b) = \(22 \to (a+b)(x+y)=22\). Thus, (a+b) CAN NEVER EQUAL ZERO.
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As a DS problem we are supposed to consider extreme cases, thus the question should have included the fact that a + b =!0.
If this was a GMAT problem the above fact would have been mentioned.

If \(x + y\) is infinity, then \(a + b\) can be equal to 0;

Lets say a = 5 b = -5 , we get a series of equations

\(x - y = 2\)
\(x - y = -2.4\)

This is a pair of parallel lines which do not intersect at any point(intersect at infinity). Hence we cannot say if \(x > y\) or NOT.

Asking Bunuel to look into this question.
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Why can't it be D ?
If a/b >1,
Then a>b.

Experts: Please explain

You cannot multiply both sides of a/b > 1 by b and write a > b, because you don't know the sign of b. If b > 0, then yes, a > b but if b < 0, then when multiplying by negative value (b), we must flip the sign and write a < b.

Hope it's clear.
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i have also added both equation
and getting (a+b)(x+y)=22.
putting any vale as a>b
say a=2 b=0 then x can be 11 and y=0 making x>y or y=11 and x=0 making X<y.
can any one tell me why its not working here????
because if i put those values again individually in ax + by = 10..its not satisfying same

thanx..

ay+bx=12, ax+by=10
a(y-x)+b(x-y)=2
(y-x)=-(x-y)=(-x+y)=(y-x);
a*-(x-y)+b(x-y)=2, -a(x-y)+b(x-y)=2, (x-y)(b-a)=2
if x-y<0, then b-a<0 and b<a [because, (neg)(neg)>0]
if x-y>0, then b-a>0 and b>a [because, (pos)(pos)>0]

(1) a > b sufic

a>b, then x>y

(2) a/b > 1 insufic

if a,b=-3,-1: -3/-1=3>1, then a<b
if a,b=3,1: 3/1=3>1, then a>b

Ans (A)

How is -a(x-y)+b(x-y)=2 =>(x-y)(a-b)=2 ?

Shouldn't it be (x-y)(b-a) = 2 ?

End result (A) is correct but the steps are wrong.