Does linear equation y=3*x + 2 contain point (r , s ) ?

[gander123]

Hey folks

here comes the question:

"In the xy-plane, does the line with equation y=3*X +2 contain the point (r,s) ?

(1) (3r + 2 - s)(4r + 9 - s) = 0

(2) (4r - 6 - s)(3r + 2 - s) = 0

The correct answer here is:[spoiler] C (both together)[/spoiler]"

My thoughts:

When I first saw this question, my first intuition was to solve it algebraically. However, this turned out to be a complete disaster and would have killed my timing. So I think there must be another short way to do it.

Well, in order to proof whether a point (r,s) is on the linear equation in slope-intercept form, I think you need the coordinates. So the values for s = ? and r = ? need to be determined. However, taking either (1) or (2) would only provide us with s in terms of r or r in terms of s. Thus taking the two independent linear equations together, one could solve for exact values for r and s and thus these values could be substituted for x and y in the equation.

Is my assumption right ?! This would not need any calculations or algebra...

Cheers,

Tobi

[aman88]

Right! I solved this problem by doing exactly the same as you mentioned. But DS problems, at times, can be deceitful and therefore, I would recommend doing some algebra to know for sure that both the statements are insufficient.

IMO C
Thanks.

[Param800]

Hey ,

I did this question like this--

QUESTION STEM :- We need to show that in the xy-plane, does the line with equation y=3*X +2 contain the point (r,s) --

So..this just means that r,s should satisfy the equation of this line ie we should prove that
s= 3r+2... which is same thing as proving (3r-s+2)= 0

NOW OPTIONS--

St. 1
(3r + 2 - s)(4r + 9 - s) = 0

We can say (3r + 2 - s)= 0 OR (4r + 9 - s) = 0

So.. we don`t have one unique solution...so insufficient.

St.2
(4r - 6 - s)(3r + 2 - s) = 0

Similarly, we can say that (4r - 6 - s)=0 OR (3r + 2 - s) = 0

So, again we don`t have any unique solution...so insufficient

COMBINATION-

When we combine them, we can for sure say that (3r + 2 - s) = 0 as it is the common solution in these two statements. And, that`s what we had to prove, so ANS = C ( both together)

P.S.

Be CAREFUL, I have seen some questions in which they trick us by giving second option like this-

(8r + 18 - 2s)(3r + 2 - s) = 0

This is exactly the first option, only difference is that I multiplied the first term by a constant(2).

So, in this case we can`t get our desired result and the answer would be E ( none of the options).

Hope this helps

Hey folks

here comes the question:

"In the xy-plane, does the line with equation y=3*X +2 contain the point (r,s) ?

(1) (3r + 2 - s)(4r + 9 - s) = 0

(2) (4r - 6 - s)(3r + 2 - s) = 0

The correct answer here is:[spoiler] C (both together)[/spoiler]"

My thoughts:

When I first saw this question, my first intuition was to solve it algebraically. However, this turned out to be a complete disaster and would have killed my timing. So I think there must be another short way to do it.

Well, in order to proof whether a point (r,s) is on the linear equation in slope-intercept form, I think you need the coordinates. So the values for s = ? and r = ? need to be determined. However, taking either (1) or (2) would only provide us with s in terms of r or r in terms of s. Thus taking the two independent linear equations together, one could solve for exact values for r and s and thus these values could be substituted for x and y in the equation.

Is my assumption right ?! This would not need any calculations or algebra...

Cheers,

Tobi

[GMATGuruNY]

Hey folks

here comes the question:

"In the xy-plane, does the line with equation y=3*X +2 contain the point (r,s) ?

(1) (3r + 2 - s)(4r + 9 - s) = 0

(2) (4r - 6 - s)(3r + 2 - s) = 0
i

If (r,s) is a point on the line y = 3x + 2, then s = 3r + 2, and 3r - s = -2. Thus, the question can be rephrased:

Does 3r - s = -2?

Statement 1: (3r+2-s)(4r+9-s) = 0
Either 3r+2-s = 0 or 4r+9-s = 0.
If 3r+2-s = 0, then 3r - s = -2.
If 4r+9-s = 0, then 4r - s = -9.
INSUFFICIENT.

Statement 2: (4r-6-s)(3r+2-s) = 0
Either 4r-6-s=0 or 3r+2-s = 0.
If 4r-6-s = 0, then 4r - s = 6.
If 3r+2-s = 0, then 3r - s = -2.
INSUFFICIENT.

Statements 1 and 2 combined:
4r - s = -9 (from statement 1) and 4r - s = 6 (from statement 2) cannot both be true, since 4r - s cannot be equal to more than one value.
Thus, the only way that the equations in the two statements can both be equal to 0 is if 3r - s = -2.
SUFFICIENT.

The correct answer is C.

[gander123]

Thanks for your contributions..great thoughts there !

closed..

cheers,

Tobi