Substitution in fraction: Given that x/y = 2/7, evaluate (7x + y) / (x - (y/7))

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chijioke

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Given that[math]\frac{ x }{ y } = \frac{ 2 }{ 7 }[/math], evaluate [math]\frac{ 7 x + y }{ x - \frac{ 1 }{ 7 } y }[/math]I choose to do it in two ways.
Way1:
I reduced the fraction by dividing the numerator and denominator by y so that we can easily obtian [math]\frac{x}{y}[/math] for the substitution to happen.
[math]\frac{ \left( 7 x + y \right) \frac{ 1 }{ y }}{ \left(x - \frac{ y }{ 7 }\right) \frac{ 1 }{ y } }=\frac{\frac{ 7 x }{ y } + \frac{ y }{ y }}{ \frac{ x }{ y } + \frac{ y }{ 7y } }[/math] [math]= \frac{ 7 \left( \frac{ x }{ y } \right) + 1}{ \frac{ x }{ y } - \frac{ 1 }{ 7 } } = \frac{ \cancel{7 } \left( \frac{ 2 }{\cancel{ 7}} \right) + 1}{ \frac{ 2 }{ 7} - \frac{ 1 }{ 7 } }[/math] [math]= \frac{\frac{ 2 + 1 }{ 2 - 1 }}{ 7 } = \frac{\frac{ 3 }{ 1 }}{ 7 }[/math] [math]= 3 \div \left( \frac{ 1 }{ 7 } \right) = 3 \times \frac{ 7 }{ 1 } = 3 \times 7 = 21[/math]
Way 2: I looked at [math]\frac{ x }{ y } = \frac{ 2 }{ 7 }[/math] as a ratio where x = 2 and y = 7. Based on that thinking, l substituted for 2 and 7 respectively for x and y directly in the expression [math]\frac{7x+y}{x-\frac{1}{7}y}[/math]
[math]\frac{7x+y}{x-\frac{1}{7}y} = \frac{7(2)+7}{2-\frac{7}{7}}=[/math][math]\frac{14+7}{2-1} = \frac{21}{1}=21[/math]The two ways gave the same solution. My question is which of the way is better and why? Or are just the two method better and depends on the choice the student wants to adopt?
 
chijioke said:
Given that[math]\frac{ x }{ y } = \frac{ 2 }{ 7 }[/math], evaluate [math]\frac{ 7 x + y }{ x - \frac{ 1 }{ 7 } y }[/math]I choose to do it in two ways.
Way1:
I reduced the fraction by dividing the numerator and denominator by y so that we can easily obtian [math]\frac{x}{y}[/math] for the substitution to happen.
[math]\frac{ \left( 7 x + y \right) \frac{ 1 }{ y }}{ \left(x - \frac{ y }{ 7 }\right) \frac{ 1 }{ y } }=\frac{\frac{ 7 x }{ y } + \frac{ y }{ y }}{ \frac{ x }{ y } + \frac{ y }{ 7y } }[/math] [math]= \frac{ 7 \left( \frac{ x }{ y } \right) + 1}{ \frac{ x }{ y } - \frac{ 1 }{ 7 } } = \frac{ \cancel{7 } \left( \frac{ 2 }{\cancel{ 7}} \right) + 1}{ \frac{ 2 }{ 7} - \frac{ 1 }{ 7 } }[/math] [math]= \frac{\frac{ 2 + 1 }{ 2 - 1 }}{ 7 } = \frac{\frac{ 3 }{ 1 }}{ 7 }[/math] [math]= 3 \div \left( \frac{ 1 }{ 7 } \right) = 3 \times \frac{ 7 }{ 1 } = 3 \times 7 = 21[/math]
Way 2: I looked at [math]\frac{ x }{ y } = \frac{ 2 }{ 7 }[/math] as a ratio where x = 2 and y = 7. Based on that thinking, l substituted for 2 and 7 respectively for x and y directly in the expression [math]\frac{7x+y}{x-\frac{1}{7}y}[/math]
[math]\frac{7x+y}{x-\frac{1}{7}y} = \frac{7(2)+7}{2-\frac{7}{7}}=[/math][math]\frac{14+7}{2-1} = \frac{21}{1}=21[/math]The two ways gave the same solution. My question is which of the way is better and why? Or are just the two method better and depends on the choice the student wants to adopt?
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Your first method actually shows that the result does not depend on specific values of x and y, but only on their ratio. You second method does not do that: we get the answer for 2 and 7, but what about 20 and 70, or any other pair with the same ratio?
 
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chijioke said:
The two ways gave the same solution. My question is which of the way is better and why? Or are just the two method better and depends on the choice the student wants to adopt?
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The second is equivalent to the first, only if you know already that the second expression depends only on x/y. So the first, which doesn't depend on knowing that, is better.

The second way could be modified by letting x=2k and y=7k; then the fact that k is fully canceled proves that the answer is unique.
 
The first method is fine.

2/7 = 4/14 = 6/21,....
You chose x=2 and y=7 and luckily got that the expression equals 21. However, if you used x=4 and y=14 you wouldn't have necessarily obtained 21 again.
 
I would write the given expression as:

\(\displaystyle \frac{7(7x+y)}{7x-y}\)

Then with:

\(\displaystyle 7x=2y\)

We have;

\(\displaystyle \frac{7(2y+y)}{2y-y}=\frac{21y}{y}=21\)
 
Steven G said:
The first method is fine.

2/7 = 4/14 = 6/21,....
You chose x=2 and y=7 and luckily got that the expression equals 21. However, if you used x=4 and y=14 you wouldn't have necessarily obtained 21 again.
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I expect not obtain 21 again because the wording has been altered to some extent. But I do believe that if I use the two methods to approach the problem, whatever solution that is obtain would be equal.
 
blamocur said:
I was trying to say the opposite, but there was a missing "not", which I've added to my post.
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I took your reply to mean that any of the two methods can be applied in solving the problem. If what I am assuming is not what you mean, please say it in clear terms.
 
chijioke said:
Why would you write it like that? What did you think?
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The given equation implies:

\(\displaystyle 7x=2y\)

And having x exclusively with coefficients of 7 makes substitution simple.
 
chijioke said:
I expect not obtain 21 again because the wording has been altered to some extent. But I do believe that if I use the two methods to approach the problem, whatever solution that is obtain would be equal.
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That is not true. Why should x and y always be the fraction in reduced form? That is 4/14 is reduced to 2/7. So x=2 and y =7, NO. It is possible that x=4 and y=7 or that x=3 and y=21 or x = 1/5 and y=7/10
 
chijioke said:
I took your reply to mean that any of the two methods can be applied in solving the problem. If what I am assuming is not what you mean, please say it in clear terms.
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Is it still not clear after my edit?
 
Steven G said:
That is not true. Why should x and y always be the fraction in reduced form? That is 4/14 is reduced to 2/7. So x=2 and y =7, NO. It is possible that x=4 and y=7 or that x=3 and y=21 or x = 1/5 and y=7/10
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I don't get you.
 
Not sure which part of my post #2 is not clear, but I'll try to rephrase it: your first method solves the problem for all x and y, while your second only solves it for x=2 and y=7.
 
blamocur said:
your first method solves the problem for all x and y, while your second only solves it for x=2 and y=7.
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I think the second method can also solve it for any values of x and y. Such as when
Steven G said:
x=4 and y=7 or that x=3 and y=21 or x = 1/5 and y=7/10
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May be I should show you.
For instance, when x=4 and y=7
We have x/y=4/7 and that [math]\frac{7x+y}{x-\frac{y}{7} }[/math]Way 1
Then [math]\frac{ \left(7x+y\right)\frac{1}{y} } { \left(x-\frac{y}{7}\right)\frac{1}{y} }=\frac{ 7\left(\frac{x}{y}\right)+\frac{y}{y} } {\frac{x}{y}-\frac{1}{7}\left(\frac{y}{y}\right) }[/math][math]= \frac{ \cancel{7}\left( \frac{4}{\cancel{7} } \right) +1 } { \frac{4}{7}- \frac{1}{7}(1)}= \frac{\frac{ 4 + 1 }{ 4 - 1 }} { 7 }[/math][math]=5÷ \frac{3}{7}= 5×\frac{7}{3}[/math][math]=\frac{35}{3} = 11\frac{2}{3}[/math]
Way 2
[math]\frac{7x+y} {x-\frac{1}{7}y}[/math] as x=4 and y =7
Then [math]\frac{7(4)+7} {4-\frac{7}{7}}[/math][math]= \frac{28+7}{4-1} = \frac{35}{3} = 11\frac{2}{3}[/math]As you can see I obtain the same solution as I use the two methods. So what else are you saying?
 
chijioke said:
As you can see I obtain the same solution as I use the two methods.
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Sorry, but I don't see that. I simply don't understand why you call the second method a solution. You are asked to solve this for any x and y for which x/y=2/7, but in your "Way 2" you "solve" it only for x=2 and y=7.

chijioke said:
So what else are you saying?
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Sorry, but I don't have anything to add.
 
blamocur said:
You are asked to solve this for any x and y for which x/y=2/7, but in your "Way 2" you "solve" it only for x=2 and y=7.
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x/y is a fraction which is translated as 2/7. Fractions can also been seen as ratio. That is x/y = x:y= 2/7 = 2:7. The terms of the ratio are x and y or 2 and 7. That is why I am applying the second method.
 
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