+0  
 
0
60
6
avatar+274 

Hi good people!,

 

please help me with this problem:

 

If:

 

\({x \over y}={2 \over 3} and {y \over z}={7 \over 5} \)

 

determine the value of \(z \over x\)

juriemagic  Oct 29, 2018
 #1
avatar+27128 
+1

Like so:

 

\(\frac{x}{y}\times\frac{y}{z}=\frac{2}{3}\times\frac{7}{5}\\\frac{x}{z}=\frac{14}{15}\\\frac{z}{x}=\frac{15}{14}\)     

Alan  Oct 29, 2018
 #2
avatar
+1

Hi Alan,

 

thanx for the response, but I do not understand..

 

Why is the result of step 1, equal to \(x \over z\) ?

 

I fail to see how the result connects to "x" and "z"...please explain if that's okay?

 

thank you kindly..

Guest Oct 29, 2018
 #3
avatar+2255 
+1

There is a common factor of y in the numerator and denominator of \(\frac{x}{\textcolor{red}{y}}*\frac{\textcolor{red}{y}}{z}\). Thus, it can be canceled out.like Alan showed. \(\frac{x}{y}*\frac{y}{z}=\frac{x}{z} \)

 

We know from the given information that \({x \over y}={2 \over 3} \text{ and } {y \over z}={7 \over 5}\). Substitute these values in for \(\frac{x}{y}\text{ and } \frac{y}{z}\). I guess I will rewrite Alan's work:

 

\(\textcolor{red}{\frac{x}{y}}*\textcolor{blue}{\frac{y}{z}}=\textcolor{green}{\frac{x}{z}}\\ \textcolor{red}{\frac{2}{3}}*\textcolor{blue}{\frac{7}{5}}=\textcolor{green}{\frac{14}{15}}\)

 

At this point, Alan just took the reciprocal of both sides to get the equation in terms of \(\frac{z}{x}\):

 

\(\frac{x}{z}=\frac{14}{15}\\ \boxed{\frac{z}{x}=\frac{15}{14}}\)

 

"I fail to see how the result connects to "x" and "z"...please explain if that's okay?"

 

The original question asked for the ratio of z to x, and we have found that answer to be 15 to 14. Does that clear it up for you?

TheXSquaredFactor  Oct 29, 2018
 #4
avatar+274 
+1

Hello guest and Alan,

 

sorry for coming back only now...I guess this is one of those sums that I just never will grasp, simply because:

 

guest stated (and Alan), I suppose, that the "y" can be canceled, because it is a common factor...okay, we can do that when we have only the variables, but when they are assigned values, it's a diferent story, because I cannot just cancel the "Y" values..so, I cannot just cancel the "7" and the "3". So we multiply with them. This gives me the result of \(15 \over14\), but this answer includes "y"..so the answer is not purely "x" and "z" only..maybe I'm over-complicating things here, but I just cannot see it.

 

It;s to say this: x * y = x, and y * z = z...which is not true

 

anycase, you guys had been helpful, I do appreciate...

juriemagic  Oct 31, 2018
edited by juriemagic  Oct 31, 2018
 #5
avatar+93866 
+1

I think you are overcomplicating things Juriemagic.

 

You can substitute any value you want (except for 0) for y and it will always cancel out.

Melody  Oct 31, 2018
 #6
avatar+93866 
0

Sets see if I can try and work out what you are saying.

 

 

You have said ... what if ??

 

x * y = x, and y * z = z  

then x=1 and y=1         

 

This does not work for the equation you are given because  x/y would equal 1 and we are told it equals 2/3

 

\({x \over y}={2 \over 3}\;\;\; and\;\;\; {y \over z}={7 \over 5}\)

Melody  Oct 31, 2018

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