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How many integral values of k are possible, if the lines 3x+

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How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 09 Mar 2011, 23:43
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How many integral values of k are possible, if the lines 3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A. 5
B. 4
C. 3
D. 6
E. 2
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Re: How many integral values of k are possible  [#permalink]

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New post Updated on: 10 Mar 2011, 00:49
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vjsharma25 wrote:
How many integral values of k are possible,if the lines
3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A) 5 B) 4 C) 3 D) 6 E) 2

I was able to do it but I am not convinced by my approach.
Please explain!!


We are given

\(3x+4ky = -6\) .... (1) and
\(kx-3y = -9\) ........(2)

Lets calculate their intersection point in terms of k

Multiply (1) by k and (2) by 3 and then subtract (2) from (1) to solve for y, we will get

\((4k^2+9)*y = 27-6k\) or \(y = (27-6k)/(4k^2+9)\).... (3)

Now, multiply (2) by 4/3k and then add (1) and (2) to solve for x, we will get

\(x = -3*(12k+6)/(4k^2+9)\).... (4)

We know that these intersection points lie in 2nd quadrant, so y has be positive and x has to be negative

y is positive when \(27-6k > 0\)or \(k < 4.5\)
Similarly, x is negative when \(12+6k > 0\) or \(k > -0.5\)

So, we have -0.5<k<4.5, so it can take integer values of 0,1,2,3,4. Hence, five values.

This is tedious approach and takes at least 2.5 to 3 minutes.

I don't know of a faster way to do so :cry:

Originally posted by beyondgmatscore on 10 Mar 2011, 00:46.
Last edited by beyondgmatscore on 10 Mar 2011, 00:49, edited 1 time in total.
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Re: How many integral values of k are possible  [#permalink]

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New post Updated on: 10 Mar 2011, 00:26
I got -0.5 < k < 4.5

hence the integral values are 0,1,2,3,4. should be B. Please check. I am not a fan of algebra, sorry :o

vjsharma25 wrote:
How many integral values of k are possible,if the lines
3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A) 5 B) 4 C) 3 D) 6 E) 2

I was able to do it but I am not convinced by my approach.
Please explain!!

Originally posted by gmat1220 on 10 Mar 2011, 00:15.
Last edited by gmat1220 on 10 Mar 2011, 00:26, edited 1 time in total.
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:24
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gmat1220 wrote:
I got -0.5 < k < 3.375

hence the integral values are 0,1,2,3. should be B

vjsharma25 wrote:
How many integral values of k are possible,if the lines
3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A) 5 B) 4 C) 3 D) 6 E) 2

I was able to do it but I am not convinced by my approach.
Please explain!!

Can you explain how you got this range?
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:32
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solving for values of x and y from original equations. And knowing x < 0 and y > 0 in II quadrant.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)

Hence -18 - 36 k < 0
So k > -0.5

And 108 - 24k >0
k < 108 / 24
k < 18 / 4
k < 9/2
k < 4.5

-0.5 < k < 4.5

Phew !
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:44
@gmat1220
But now you got the correct range and correct answer :-).

I was not taking values of x and y as independent. Thats why i got confused.
Thanks for the explanation.
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Re: How many integral values of k are possible  [#permalink]

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New post 10 Mar 2011, 00:48
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Man this was monster ! especially my equations go completely haywire when I look at algebra.
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 29 Jun 2015, 22:49
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Can someone pls explain how to arrive at these 2 equations mentioned above.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 29 Jun 2015, 23:34
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jayanthjanardhan wrote:
Can someone pls explain how to arrive at these 2 equations mentioned above.
x = -18 - 36k / (4k^2 + 9)
y = (108 - 24k) / 4 (4k^2 + 9)



The given equations are

3x + 4ky = -6 ------1
kx - 3y = -9 ------2

Try solving the two equations:

1. Multiply eq 1 by k, 3kx + 4k^2y = -6k ---------3
2. Multiply eq 2 by 3, 3kx - 9y = -27 ---------4

Solving eq, 3 and 4, we can get the value of y.

where y = (27-6k)/(4k^2 + 9)..

Since y should be > 0, 27-6k > 0, solving k<4.5

Similarly, solving for x gives or sub y in eq 2 , x = -6(6k+3) / 4k^2 + 9

since x < 0, 6k+3 should be > 0 => 6k+3 > 0, solving k < -0.5

Hence k has 5 Integral values 0, 1, 2, 3 and 4.
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 13 Dec 2018, 02:24
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I have a question..
when we get- y=27-6k/4k^2+9 and since y>0 then why do we assume just- 27-6k>0 instead of 27-6k>0 and 4k^2+9>0 OR 27-6k<0 and 4k^2+9<0
I know I'm missing something here but I can't understand what..Please help!
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 07 Jan 2019, 09:32
The value 4k^2+9>0 will always be positive its the numerator which will make the diff

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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 19 Jan 2019, 08:16
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vjsharma25 wrote:
How many integral values of k are possible, if the lines 3x+4ky+6 = 0, and kx-3y+9 = 0 intersect in the 2nd quadrant.

A. 5
B. 4
C. 3
D. 6
E. 2


The solution provided by beyondgmatscore is perfect. A student asked me to add a few steps, so that what I'll do.

GIVEN:
3x + 4ky + 6 = 0
kx - 3y + 9 = 0

Rewrite as:
3x + 4ky = -6
kx - 3y = -9

NOTE: The intersection of the two lines will be at the point where the same pair of x- and y-values satisfy BOTH equations.
In other words, we want to SOLVE the system of equations above.

So, let's use the ELIMINATION method to solve the system.

First we'll eliminate the x terms
So, multiply both sides of the TOP equation by k, and multiply both sides of the BOTTOM equation by 3 to get:
3kx + 4k²y = -6k
3kx - 9y = -27

Subtract the bottom equation from the top equation to get: 4k²y + 9y = -6k + 27
Rewrite as: y(4k² + 9) = 27 - 6k
Divide both sides by (4k² + 9) to get: y = (27 - 6k)/(4k² + 9)

KEY CONCEPT: If the two lines intersect in the 2nd quadrant, then the x-value must be NEGATIVE, and the y-value must be POSITIVE

If the y-value must be POSITIVE, we can write: (27 - 6k)/(4k² + 9) > 0
Since (4k² + 9) is POSITIVE for all values of k, we can conclude that 27 - 6k is POSITIVE
In other words: 27 - 6k > 0
Add 6k to both sides to get: 27 > 6k
Divide both sides by 6 to get: 4.5 > k
--------------------------------------

Now we'll perform similar steps to isolate x (by eliminating the y terms)

Take:
3x + 4ky = -6
kx - 3y = -9

Multiply both sides of the TOP equation by 3, and multiply both sides of the BOTTOM equation by 4k to get:
9x + 12ky = -18
4k²x - 12ky = -36k

Add the two equations to get: 9x + 4k²x = -18 - 36k
Factor left side: x(9 + 4k²) = -18 - 36k
Divide both sides by (9 + 4k²) to get: x = (-18 - 36k)/(9 + 4k²)

KEY CONCEPT: If the two lines intersect in the 2nd quadrant, then the x-value must be NEGATIVE, and the y-value must be POSITIVE

If the x-value must be NEGATIVE, we can write: (-18 - 36k)/(9 + 4k²) < 0
Since (9 + 4k²) is POSITIVE for all values of k, we can conclude that -18 - 36k is NEGATIVE
In other words: -18 - 36k < 0
Add 18 to both sides to get: -36k < 18
Divide both sides by -36 to get: k > -0.5
--------------------------------------

Now combine both results to get: -0.5 < k < 45

How many integral values of k are possible?
If -0.5 < k < 45, then the possible INTEGER values of k are: 0, 1, 2, 3, 4 (5 possible values)

Answer: A

Cheers,
Brent
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Re: How many integral values of k are possible, if the lines 3x+  [#permalink]

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New post 01 Feb 2019, 23:07
solve both quadratic equations
x = (6k+ 1)(-6) / (4k2 +9)

y = (27-6k) / (4k2 +9)

He says that they intersect in 2nd quadrant.

Hence, x < 0, y >0 ( Any value in second quadrant)
x <0
-6(1+6k) < 0
k> -1/6
K>-0.166 ----1

y>0
27-6k > 0
6k < 27
k < 27/6
k < 4.5 ---2

Hence from 1 & 2
-0.166< K < 4.5

Integral values K can take = 0,1,2,3,4
Answer is 5 option A.
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Re: How many integral values of k are possible, if the lines 3x+   [#permalink] 01 Feb 2019, 23:07
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