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DS "distinct linear equations rule" - exception?

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by Testtrainer » Fri Jul 09, 2010 8:53 pm
What is the value of x?

(1) 3x = 2y

(2) 5y = 3z and 5x = 2z

The answer is E! I checked the math, and its true - you can't solve for any variable. However, aren't these equations distinct and linear? What rule of math says that I can't solve? Thanks for any help.
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Source: — Data Sufficiency |

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by Rahul@gurome » Fri Jul 09, 2010 9:10 pm
What is the value of x?

(1) 3x = 2y

(2) 5y = 3z and 5x = 2z

The answer is E! I checked the math, and its true - you can't solve for any variable. However, aren't these equations distinct and linear? What rule of math says that I can't solve? Thanks for any help.

(1) 3x = 2y implies x = 2y/3, but we don't know the value of y.
So, (1) is NOT SUFFICIENT.

(2) 5y = 3z and 5x = 2z implies x = 2z/5 = (2/5)(5y/3) = 2y/3, but we don't know the value of y.
So, (2) is NOT SUFFICIENT.

Combining (1) and (2), we get no solution, as the equations from (1) and (2) are the same and not distinct.

The correct answer is (E).

Does that help?
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by Testtrainer » Fri Jul 09, 2010 9:26 pm
Thanks for the super-quick reply. I did indeed see what you worked out. I guess my question is: how can I learn to recognize that such seemingly distinct equations are actually not? Otherwise, I might be tempted to work out every single DS equation I come across (which I know I shouldn't do), just in case I get caught again...
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by Rahul@gurome » Fri Jul 09, 2010 9:34 pm
Testtrainer wrote:Thanks for the super-quick reply. I did indeed see what you worked out. I guess my question is: how can I learn to recognize that such seemingly distinct equations are actually not? Otherwise, I might be tempted to work out every single DS equation I come across (which I know I shouldn't do), just in case I get caught again...
You can solve them for one variable, for example, here we wanted to find x, so we solved x in terms of other variables, but from both the statements we get the same equation, x = 2y/3, which cannot be solved further. Solving one variable in terms of other helps to know whether the equation can be solved or not. Does that answer your question?
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by Testtrainer » Fri Jul 09, 2010 9:51 pm
So sorry, but not quite.

Here's the deal: whenever I see multiple equations with multiple variables, I run through various "checks": are they linear? are they distinct? is there another way to solve? Typically, I can see quite quickly whether the equations are distinct. In this particular case, however, they appeared (clearly) to be distinct.

My question: what tool can I use to quickly recognize whether the equations are distinct?

After all, I can't work through every set of equations that I see. I have a lot of other math to do...
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by Testtrainer » Sat Jul 10, 2010 9:59 pm
So I think I might have answered my own question, but I could use some confirmation. The following equations:
3x = 2y, 5y = 3z, and 5x = 2z are indeed 3 distinct linear equations with 3 unknowns. However, each linear equation has a y-intercept of 0, meaning that all 3 equations intersect through (0,0). Since they don't intersect anywhere else, they can't have any common solutions.

If my reasoning is correct, this means that no more than one equation can have a y-intercept of 0. So in checking whether equations are distinct and linear, I think we have to make sure that no more than one equation has a y-intercept of 0. Equations with a y-intercept of 0 have no addition or subtraction of stand-alone values. For example, 2x = 3y has a y-intercept of 0, while 2x + 6 = 3y has a y-intercept of 2 (6 divided by 3).

Am I on the right track here? I've never heard of anything like this before, so I feel like I'm missing something...
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by Rahul@gurome » Sun Jul 11, 2010 4:26 am
Testtrainer wrote:So I think I might have answered my own question, but I could use some confirmation. The following equations:
3x = 2y, 5y = 3z, and 5x = 2z are indeed 3 distinct linear equations with 3 unknowns. However, each linear equation has a y-intercept of 0, meaning that all 3 equations intersect through (0,0). Since they don't intersect anywhere else, they can't have any common solutions.

If my reasoning is correct, this means that no more than one equation can have a y-intercept of 0. So in checking whether equations are distinct and linear, I think we have to make sure that no more than one equation has a y-intercept of 0. Equations with a y-intercept of 0 have no addition or subtraction of stand-alone values. For example, 2x = 3y has a y-intercept of 0, while 2x + 6 = 3y has a y-intercept of 2 (6 divided by 3).

Am I on the right track here? I've never heard of anything like this before, so I feel like I'm missing something...
That's right!
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by sayanpaul » Tue Apr 09, 2013 9:50 pm
(1) 3x = 2y. We cannot solve for x since there are 2 variables.
(2) 5y-3z =0 and 5x-2z =0
Solving these two equations we get, 10y = 15x or 2y = 3x
Since both the equations in (1) and (2) are equivalent, we cannot solve for x.
Hence E
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by GMATGuruNY » Wed Apr 10, 2013 7:25 am
Testtrainer wrote:What is the value of x?

(1) 3x = 2y

(2) 5y = 3z and 5x = 2z
Just to clarify: when the statements are combined, we have 3 variables but only TWO distinct linear questions.
Statement 2:
5y = 3z --> 10y = 6z.
5x = 2z --> 15x = 6z.

Linking together 10y = 6z and 15x = 6z, we get:
10y = 6z = 15x
10y = 15x
2y = 3x.
This is the equation given in statement 1.
Thus, statement 1 DOES NOT provide a third equation.

The following cases satisfy both statements:
If 10y = 6z = 15x = 30, then y=3, z=5, and x=2.
If 10y = 6z = 15x = 60, then y=6, z=10, and x=4.
Since x can be different values, the two statements combined are INSUFFICIENT.

The correct answer is E.
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