+0

0
254
6
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Hi good people!,

If:

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

determine the value of $$z \over x$$

Oct 29, 2018

#1
+27850
+2

Like so:

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

Oct 29, 2018
#2
+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
+2338
+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
+448
+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

juriemagic  Oct 31, 2018
edited by juriemagic  Oct 31, 2018
#5
+101044
+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
+101044
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