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kapilnegi
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I think C.

(3r+2-s)(4r+9-s)=0 -> 3r+2-s=0 OR 4r+9-s=0
(4r-6-s)(3r+2-s)=0 -> 4r-6-s=0 OR 3r+2-s=0

which gives 3r+2-s=0 or s=3r+2
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goalsnr
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x97agarwal
In the xy-plane, does the line with equation y=3x+2 contain (r,s)?

1. (3r+2-s)(4r+9-s)=0; = 13r^2 +35 r - 7rs + 8r - 11s + s^2 + 36
2. (4r-6-s)(3r+2-s)=0

S1. Tells you that (3r+2-s) = 0; s = 3r + 2
S2. Tells you that (3r+2-s) = 0; s = 3r + 2

so (3r+2-s) is the common value in both the equations and this satisfies the equation y = 3x + 2

IMO C
X97, Stat1 and 2 give the same info. I dont understand why C? Pls explain.
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when x*y = 0

the information we have is either x=0 OR y=0

statement 1 gives either (3r+2-s) = 0 OR (4r+9-s)=0. so the answer of our question could be either "yes" or "no"....

Similiar logic for statement 2.

thats why C is the answer.

its something like this

Question x = ?
1. x=3 or 4
2. x=3 or 5 Option C
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x97agarwal
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goalsnr
x97agarwal
In the xy-plane, does the line with equation y=3x+2 contain (r,s)?

1. (3r+2-s)(4r+9-s)=0;
2. (4r-6-s)(3r+2-s)=0

S1. Tells you that (3r+2-s) = 0; s = 3r + 2
S2. Tells you that (3r+2-s) = 0; s = 3r + 2

so (3r+2-s) is the common value in both the equations and this satisfies the equation y = 3x + 2

IMO C
X97, Stat1 and 2 give the same info. I dont understand why C? Pls explain.

From S1. (3r+2-s) = 0 or (4r+9-s)=0
S2. (4r-6-s) = 0 or (3r+2-s)=0
Exaple: If xy = 0 then either x = 0 or y = 0 or x and y both = 0
Similarily:
Lets assume that (3r+2-s) is not = 0 and (4r-6-s) = 0 and (4r+9-s)=0
Now equate both these equations, (4r-6-s) = (4r+9-s), you will get 15 = 0, which is not possible

The only way the equation can have a solution is if (3r+2-s) = 0
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goalsnr
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x97agarwal
goalsnr
x97agarwal
In the xy-plane, does the line with equation y=3x+2 contain (r,s)?

1. (3r+2-s)(4r+9-s)=0;
2. (4r-6-s)(3r+2-s)=0

S1. Tells you that (3r+2-s) = 0; s = 3r + 2
S2. Tells you that (3r+2-s) = 0; s = 3r + 2

so (3r+2-s) is the common value in both the equations and this satisfies the equation y = 3x + 2

IMO C
X97, Stat1 and 2 give the same info. I dont understand why C? Pls explain.

From S1. (3r+2-s) = 0 or (4r+9-s)=0
S2. (4r-6-s) = 0 or (3r+2-s)=0
Exaple: If xy = 0 then either x = 0 or y = 0 or x and y both = 0
Similarily:
Lets assume that (3r+2-s) is not = 0 and (4r-6-s) = 0 and (4r+9-s)=0
Now equate both these equations, (4r-6-s) = (4r+9-s), you will get 15 = 0, which is not possible

The only way the equation can have a solution is if (3r+2-s) = 0


Ah...Get it. 4 years of engineering were just a waste of time :(

Thanks. Kudos.



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