hello guys got stuck on my last exercise of ratio today, Is kind of easy but i don't know what steps to go for... i did read the book but im still confused

http://s2.postimage.org/14mwmbwro/Ratio_3.jpg

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- Jan 20th 2011, 05:13 PMvaironxxrdRatio 3 problem medium
hello guys got stuck on my last exercise of ratio today, Is kind of easy but i don't know what steps to go for... i did read the book but im still confused

http://s2.postimage.org/14mwmbwro/Ratio_3.jpg - Jan 20th 2011, 05:20 PMdwsmith
- Jan 21st 2011, 07:03 PMvaironxxrd
- Jan 21st 2011, 07:05 PMdwsmith
- Jan 21st 2011, 07:13 PMvaironxxrd
- Jan 21st 2011, 07:18 PMdwsmith
$\displaystyle \displaystyle\frac{2}{x}=\frac{7}{y}\cdot\frac{7}{ 7}$

7/7 is 1, so we aren't doing anything to the equation by multiplying by 1.

$\displaystyle \displaystyle\frac{1}{7}\cdot\frac{2}{x}=\frac{7}{ y}\cdot\frac{1}{7}\Rightarrow\frac{2}{7x}=\frac{1} {y}$

$\displaystyle \displaystyle\frac{x}{x}\cdot\frac{2}{7x}=\frac{1} {y}\Rightarrow x\cdot\frac{2}{7x}=x\cdot\frac{1}{y}$

x/x is 1 too so we didn't do anything to the equation.

$\displaystyle \displaystyle\frac{2}{7}=\frac{x}{y}$ - Jan 21st 2011, 07:45 PMvaironxxrdRation
OH that is the answer? Still confusing for me :(! Oh i see some notes on this but it just does not match the problem

- Jan 21st 2011, 07:49 PMdwsmith
- Jan 21st 2011, 07:59 PMvaironxxrd
- Jan 21st 2011, 08:01 PMdwsmith
- Jan 21st 2011, 08:01 PMvaironxxrd
- Jan 21st 2011, 08:08 PMdwsmith
$\displaystyle \displaystyle\frac{2}{x}=\frac{7}{y}\cdot\frac{7}{ 7} \ \text{7 divided by 7 is 1 so we aren't changing the equality by multiplying by 1.}$

$\displaystyle \displaystyle\frac{1}{7}\cdot\frac{2}{x}=\frac{7}{ y}\cdot\frac{1}{7} \ \text{lets prove this is the same as the above equation.}$

If we multiply both sides by 7, we will have 1 and 7/7.

$\displaystyle \displaystyle 7\cdot\frac{1}{7}\Rightarrow\frac{7}{7}\Rightarrow 1\cdot\frac{2}{x}=\frac{7}{y}\cdot\frac{1}{7}\cdot 7\Rightarrow\frac{2}{y}\cdot\frac{7}{7} \ \text{as you can see, the 7 on the left hand side back to the right.}$

Same for the x. - Jan 21st 2011, 08:15 PMvaironxxrd