Equation

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Equation

by seema19 » Thu Oct 06, 2011 4:52 am
What is the value of k?

1. The system of equation,
2x + 3y = 7
3x + ky = 15
has a unique solution

2. The system of equations,
2x + 3y = 7
3x + ky = 21/2
has infinite number of solutions.

Answer: B

Can somebody please explain the solution in detail.
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by Brent@GMATPrepNow » Thu Oct 06, 2011 5:43 am
seema19 wrote:What is the value of k?

1. The system of equation,
2x + 3y = 7
3x + ky = 15
has a unique solution

2. The system of equations,
2x + 3y = 7
3x + ky = 21/2
has infinite number of solutions.

Answer: B

Can somebody please explain the solution in detail.
For a system of 2 equations and 2 variables to have an infinite number of solutions, it must be the case that the 2 equations are equivalent.

Example:
The system
3x+2y=7
6x+4y=14 has an infinite number of solutions since these 2 equations are equivalent. If you multiply the top equation by 2, we get 6x+4y=14, which is the bottom equation.

Statement 1:
If that system has a unique solution, then there are many possible values for k
INSUFFICIENT

Statement 2:
If that system has an infinite number of solutions, then the two equations are equivalent.
When we multiply 2x (in the top equation) by 3/2 we get 3x (in the bottom equation)
When we multiply 7 (in the top equation) by 3/2 we get 21/2 (in the bottom equation)
Since the equations are equivalent, it must be the case that when we multiply 3y (in the top equation) by 3/2 we get ky (in the bottom equation)
So, (3y)(3/2)=ky, which means (9/2)y=ky, which means k = 9/2
SUFFICIENT

The answer is B

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by GmatMathPro » Thu Oct 06, 2011 5:56 am
For a system of two equations to have infinitely many solutions, one equation should be a multiple of the other equation. Suppose you have 1 equation, x+y=8. Clearly there are infinitely many solutions to this equation because there are many (x,y) pairs that satisfy the equation: (1,7);(2,6);(3,5)... and so on. To have a unique (x,y) solution, we would need one more equation that imposes a different set of restrictions. For example, If x+y=8 and x-y=10, only (9,-1) solves both equations simultaneously. However, if the second equation is just a multiple of the first, like 2x+2y=16, then that's really just repeating the first equation; we could divide out a 2 from both sides and we wouldn't really have any new information about how x and y relate to each other, and we would still have infinitely many solutions.

The other possibility is that there are no solutions which would happen when the coefficients of x and y are in the same ratio, but the constant terms are in a different ratio. For example,

x+2y=8
2x+4y=17

Note that the coefficients of x and y are both in the ratio 1:2 whereas the constant terms are in the ratio 8:17.

Statement 1: As long as k is not equal to 4.5, this system will have a unique solution. INSUFFICIENT.

Statement 2: k=4.5 is the only value that makes the second equation a multiple of the first. That is, if k=4.5, the second equation is just the first equation times 1.5, which would make the number of solutions infinite. SUFFICIENT

Ans: B
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by MrCleantek » Wed Oct 12, 2011 10:04 am
Guys, I have a question to find out a easier way to find the roots for this equation.
n^2+n-240=0
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by GmatMathPro » Wed Oct 12, 2011 10:44 am
Easier than what?
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by MrCleantek » Wed Oct 12, 2011 10:50 am
easier than n-16 and n+15. Finding out 16 x 15 = 240. Any short cuts?
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by GmatMathPro » Wed Oct 12, 2011 11:49 am
Hmmmm.....nothing obviously superior that I'm aware of. You could use the quadratic equation. The advantage of that is that it eliminates guesswork. The disadvantage is it's tedious and you have to know the square root of 961.

How did you end up getting 15*16=240? Guessing and checking? If so, there is a way to improve on that, but first some background info to make sure we're on the same page. This is going to be kind of long, but once you see the trick you'll see it's a lot faster than guessing and checking.

If we list the factors of some perfect square, let's say 36, in pairs:

1,36
2,18
3,12
4,9
6,6

Note that in each pair, one is always less than the square root of 36 and one is always greater until we get to 6,6 where they are both exactly equal to the square root of 36. As we move from the top of the list to the bottom, the number on the left is approaching the square root of 36 from the bottom and the number on the right is approaching the square root of 36 from the top. But except for 6 and 6, there is always one number greater than 6 and one number less than 6 in each pair.

Now let's look at an integer that is not a perfect square, 72:

1, 72
2, 36
3, 24
4, 18
6, 12
8, 9

For a reference point, the square root of 72 is approximately 8.49. Note that the same thing is happening: one factor is always less than 8.49 and one is always greater than 8.49. Now, 72 has two factors that differ by 1: 8*9. Note that this is the ONLY possible pair that could have worked. The square root is about 8.49, so if 72 were going to have two factors that differ by 1, it would HAVE to be 8 and 9 to keep with the rule of each pair of factors having one number greater than the square root and one number less than the square root.

Let's say we want to find out if some other number has factors that differ by 1. If the number were 87, we could reason as follows: 9^2=81 and 10^2=100. Therefore the square root of 87 must be between 9 and 10. That is, 9 point something. So, if 87 DOES have factors that differ by 1 they must be 9 and 10. However, 9*10=90, not 87, so that must mean 87 does NOT have two factors that differ by 1.

So if we had to factor n^2-n-87, we would know for sure that it is impossible, because we've just proven that 87 doesn't have a pair of factors that differ by 1.

Now, in the case of 240, 15^2=225 and 16^2=256, so the square root of 240 must be between 15 and 16. So if 240 CAN be factored as two integers that differ by 1, it must be 15 and 16. Then you just have to verify that they work by multiplying them together.

So, to summarize, if an integer, n, does have factors that multiply to n and have a difference of 1, one of them must be first integer above the square root of n and one must be the first integer below the square root of n.
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by MrCleantek » Wed Oct 12, 2011 7:47 pm
Got it.
Thanks man.
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