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?
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?