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by yellowho » Wed Jan 19, 2011 11:18 pm
In the coordinate plane, line A is defined by the equation 3x+2y=k, and line B is defined by the
equation jx-y=-7. At what point do the two lines intersect?

(1) Line A passes through the point (0,7)
(2) Line B passes through the point (3,1)

Can someone explain why it's A? You still don't know j
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by Rahul@gurome » Wed Jan 19, 2011 11:46 pm
(1) Since line A passes through (0, 7). So 3(0) + 2(7) = k implies k = 14.
Adding 3x + 2y = 14 and 2jx - 2y = -14, we get (3 + 2j)x = 0, which implies x = 0 and y = 7
So, the two lines intersect at (0, 7). SUFFICIENT.

(2) Since line B passes through the point (3,1), so 3j - 1 = -7 implies j = -2
Adding 3x + 3y = k and -6x - 3y = -21, we get -3x = k - 21, which cannot be solved.
Hence, NOT SUFFICIENT.

The correct answer is A.

Does that help?
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by yellowho » Wed Jan 19, 2011 11:54 pm
What's your rationale for adding? I know it works fine but, I didn't "See" the reason to add since you can't eliminate the variable. What's the key takeaway here? The answer is obvious now but only because j cannot be -3/2. I guess, can you walk me through your thought process and why you knew that adding together would work? Traditional algebra would indicate that you can't eliminate j so I didn't even bother adding.
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by Rahul@gurome » Thu Jan 20, 2011 12:01 am
yellowho wrote:What's your rationale for adding? I know it works fine but, I didn't "See" the reason to add since you can't eliminate the variable. What's the key takeaway here? The answer is obvious now but only because j cannot be -3/2. I guess, can you walk me through your thought process and why you knew that adding together would work? Traditional algebra would indicate that you can't eliminate j so I didn't even bother adding.
When we look at the 2 equations, we can observe that by adding the 2 equations the 'y' term and the 'constant term' are cancelled. This is the main reason we added the equations here.
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by yellowho » Thu Jan 20, 2011 12:11 am
But you still have j and x. Algebraically, its not that helpful. It's only helpful if you know that j cannot be the same as the slope in the original line. How did you solve it? Did you say (3+2j)x=0 X= 0 or 3+2j=0 thus J=-3/2 since j cannot be -3/2 then x must be zero. Or did you some how know based on some property. I guess how did you avoid falling into the C trap?. I was thinking of combining originally by saw 3 variables and 2 equation and stopped.
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by Rahul@gurome » Thu Jan 20, 2011 1:53 am
yellowho wrote:I was thinking of combining originally by saw 3 variables and 2 equation and stopped.
The "rules" of algebras like "more unknown than equations can't be solved" etc are applicable only when there is n other information is available. When we have other data, why don't we try to use them rather than blindly following the rules?
yellowho wrote:Algebraically, its not that helpful. It's only helpful if you know that j cannot be the same as the slope in the original line. How did you solve it? Did you say (3+2j)x=0 X= 0 or 3+2j=0 thus J=-3/2 since j cannot be -3/2 then x must be zero. Or did you some how know based on some property.
If you take j = -3/2, then the equation of two lines will be same and they will coincide on each other, i.e. the two lines will intersect at infinite number of points. Thus from statement 1 we conclude either the lines intersect at point (0, 7) or at infinite number of points which includes point (0, 7) too. Hence in any case they are intersecting at point (0, 7).

Here is another approach you may find useful:
  • The Equation for line B is jx - y = -7
    Rearranging, y = jx + 7
    Thus, only the slope of the line B depends on j.
    But we know the y-intercept of line B, which is equal to 7.
    Hence B always passes through (0, 7) and we conclude it from the question stem itself.

    Now statement 1 is clearly sufficient.
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