# Ratio 3 problem medium

• January 20th 2011, 06:13 PM
vaironxxrd
Ratio 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
• January 20th 2011, 06:20 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
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

$\displaystyle\frac{1}{7}\cdot\frac{2}{x}=\frac{7}{ y}\cdot\frac{1}{7}\Rightarrow x\cdot\frac{2}{7x}=\frac{1}{y}\cdot x=\cdots$
• January 21st 2011, 08:03 PM
vaironxxrd
Quote:

Originally Posted by dwsmith
$\displaystyle\frac{1}{7}\cdot\frac{2}{x}=\frac{7}{ y}\cdot\frac{1}{7}\Rightarrow x\cdot\frac{2}{7x}=\frac{1}{y}\cdot x=\cdots$

How did you got that [Math]\displaystyle\frac{1}{7}[/tex]
The book Does not give me a 1
• January 21st 2011, 08:05 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
How did you got that [Math]\displaystyle\frac{1}{7}[/tex]
The book Does not give me a 1

Divide by 7.
• January 21st 2011, 08:13 PM
vaironxxrd
Quote:

Originally Posted by dwsmith
Divide by 7.

Sorry if ill keep asking the same question but is there any reason for this?
i mean i got [Math]\frac{2}{7}\frac{?}{?}[/tex]
So the equals sign say that we don't know that number so i believe it can be any number, You understand how im i getting confuse?
• January 21st 2011, 08:18 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
Sorry if ill keep asking the same question but is there any reason for this?
i mean i got [Math]\frac{2}{7}\frac{?}{?}[/tex]
So the equals sign say that we don't know that number so i believe it can be any number, You understand how im i getting confuse?

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

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

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

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

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

$\displaystyle\frac{2}{7}=\frac{x}{y}$
• January 21st 2011, 08:45 PM
vaironxxrd
Ration
OH that is the answer? Still confusing for me :(! Oh i see some notes on this but it just does not match the problem
• January 21st 2011, 08:49 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
OH that is the answer? Still confusing for me :(! Oh i see some notes on this but it just does not match the problem

$\displaystyle\frac{7}{7}=1$

Multiplying by doesn't change the equality. Same applies to x/x.

I don't understand what you don't get. Can you be more specific?
• January 21st 2011, 08:59 PM
vaironxxrd
Quote:

Originally Posted by dwsmith
$\displaystyle\frac{7}{7}=1$

Multiplying by doesn't change the equality. Same applies to x/x.

I don't understand what you don't get. Can you be more specific?

Ok.. Sorry for that let me try to explain.

I started with [Math]\frac{2}{X} = \frac{7}{y} Then \frac {2}{7} = \frac {?}{?}[/tex]
• January 21st 2011, 09:01 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
Ok.. Sorry for that let me try to explain.

I started with [Math]\frac{2}{X} = \frac{7}{y} Then \frac {2}{7} = \frac {?}{?}[/tex]

Then Post 6 shows the math progression.
• January 21st 2011, 09:01 PM
vaironxxrd
Quote:

Originally Posted by vaironxxrd
Ok.. Sorry for that let me try to explain.

I started with [Math]\frac{2}{X} = \frac{7}{y} Then \frac {2}{7} = \frac {?}{?}[/tex]

so i believe THe result would be {2y}{7x}[/tex] because we are multiplying it?
• January 21st 2011, 09:08 PM
dwsmith
Quote:

Originally Posted by vaironxxrd
so i believe THe result would be {2y}{7x}[/tex] because we are multiplying it?

$\displaystyle\frac{2}{x}=\frac{7}{y}\cdot\frac{7}{ 7} \ \text{7 divided by 7 is 1 so we aren't changing the equality by multiplying by 1.}$

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

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

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

Same for the x.
• January 21st 2011, 09:15 PM
vaironxxrd
Quote:

Originally Posted by dwsmith
$\displaystyle\frac{2}{x}=\frac{7}{y}\cdot\frac{7}{ 7} \ \text{7 divided by 7 is 1 so we aren't changing the equality by multiplying by 1.}$

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

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

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

Same for the x.

oh ok thanks