In the XY plane, does the line with equation y = 3x + 2

[Harish Dorai]

In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

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

[anadi]

In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

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

In case 1, one of 2 brackets can be 0. If 3r+2-s = 0 then yes, line contains the point. Is 2nd bracket value is 0 , there is one valuse of r and s on which it does, but not for each value of r and s.

Same for case 2.

Coombined together, if both are true, 4r+9-s and 4r-6-S, can not both be 0 at the same time. So 3r+2-S = 0. This is same as y=3x+2. So if 3r+2-s = 0, this line will contain every (R,S) satisfying this equation.

So C is the answer. What is OA?

[Harish Dorai]

(C) is the right answer. Great explanation!

[Guest]

Could you please elaborate or clarify by what you mean when you say, "Combined together, if both are true, 4r+9-s and 4r-6-S, cannot both be 0 at the same time." Why is this so?

Thanks

[guest612]

yes, can you please further explain the steps? for example, why are there only unique values for one of the brackets but not both for statements 1 & 2? also, isn't 3r+2-s a reiteration of the equation y=3x+2? but not sure how that answers the question and concludes that the line contains the coordinates (r,s). in sum, if you can just explain this a little more thoroughly it would be greatly appreciated! thanks.

[Guest813]

("4r-s"+9) and ("4r-s"-6) can not both be Zero...

[rfernandez]

Great solution, anadi. I'll give it a go, too.

In the XY plane, does the line with equation y = 3x + 2 contain the point (r,s)?

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

Question: Does s = 3r + 2? Or, put another way, does 3r +2 - s = 0?

(1) Each of these factors may equal zero, so we know that either 3r + 2 - s = 0 OR that 4r + 9 - s = 0. Insufficient.
(2) Similarly, we know that either 4r - 6 - s = 0 OR that 3r + 2 - s = 0. Insufficient.
(1&2) The only possibility that would satisfy both statements is that 3r + 2 - s = 0. Sufficient.

[fighting_cax]

Hi Rey,
Just to clarify: how does finding out that 3r + 2 - s = 0 answer the question of whether y = 3x + 2 contain the point (r, s)?

Thanks.

[RonPurewal]

Hi Rey,
Just to clarify: how does finding out that 3r + 2 - s = 0 answer the question of whether y = 3x + 2 contain the point (r, s)?

Thanks.

the question of whether y = 3x + 2 contains (r, s) is the same as the question of whether s = 3r + 2. this is true of equations and graphs in general: "point (this, that) is on the graph of this equation" is the same as "when you plug this and that into the equation, the equation is true".

subtract s from both sides of 3r + 2 = s, giving 3r + 2 - s = 0. they're the same equation.

[amc08]

cant we get 2 equations with 2 variables from (1) and then the values of r and s and see if it fits the equeation y= 3x+2 and the same we can do for (2) and then (1) and (2) are each sufficient and the answer is then D

??

[RonPurewal]

cant we get 2 equations with 2 variables from (1) and then the values of r and s and see if it fits the equeation y= 3x+2 and the same we can do for (2) and then (1) and (2) are each sufficient and the answer is then D

??

no.

the "two variables / two equations" approach only works when the equations are simultaneous: i.e., when they BOTH have to be true.
this isn't the case here. if (3r + 2 - s)(4r + 9 - s) = 0, that means (3r + 2 - s) = 0 OR (4r + 9 - s) = 0. therefore, 3r - s = -2 OR 4r - s = -9, which has A LOT of solutions (anything that solves either of those equations qualifies as a solution).

--

Could you please elaborate or clarify by what you mean when you say, "Combined together, if both are true, 4r+9-s and 4r-6-S, cannot both be 0 at the same time." Why is this so?

4r + 9 - s is (4r - 6 - s) plus 15.
if some number is zero, then that number plus 15 isn't zero.

[itstimdy]

Since the question asked if a point is on a line, I rephrased the question into, "Do we have enough information to find (r,s)?" Whether it falls on the line or not, it doesn't matter.

1) 1 equation 2 variables - NS
2) 1 equation 2 variables - NS

Together - after making sure they weren't the same equation, 2 variables 2 equations - S
Choice C.

I may have totally over simplified it.

[RonPurewal]

Since the question asked if a point is on a line, I rephrased the question into, "Do we have enough information to find (r,s)?" Whether it falls on the line or not, it doesn't matter.

1) 1 equation 2 variables - NS
2) 1 equation 2 variables - NS

Together - after making sure they weren't the same equation, 2 variables 2 equations - S
Choice C.

I may have totally over simplified it.

yep, you totally oversimplified it. this is one of the few problems on which that kind of approach is a lucky guess -- it's usually a trap answer.

for a half-hour or more on this problem -- including a couple of other variations that i made, on which the answer is not C -- watch the MARCH 17, 2011 lecture here:
http://www.manhattangmat.com/thursdays-with-ron.cfm