Prove that these three ratios are equivalent

[RChristenk]

Homework Statement

If $\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\dfrac{x}{y}$

Relevant Equations

Basic fraction and algebra manipulation

I have the solution:

$\dfrac{2x-3y}{3z+y} = \dfrac{z-y}{z-x} = \dfrac {x+3z}{2y-3x}$ (1)

$\dfrac{2x-3y}{3z+y} = \dfrac{3(z-y)}{3(z-x)} = \dfrac {x+3z}{2y-3x}$ (2)

$= \dfrac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)}$ (3)

$=\dfrac{x}{y}$

My question is how is it possible to go from (2) to (3). Because in (2) there are three fractions and two equal signs, yet they are merged in (3). I don't understand how this can be done. Thank you.

 

[Hornbein]

Homework Statement: If $\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\dfrac{x}{y}$
Relevant Equations: Basic fraction and algebra manipulation

I have the solution:

$\dfrac{2x-3y}{3z+y} = \dfrac{z-y}{z-x} = \dfrac {x+3z}{2y-3x}$ (1)

$\dfrac{2x-3y}{3z+y} = \dfrac{3(z-y)}{3(z-x)} = \dfrac {x+3z}{2y-3x}$ (2)

$= \dfrac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)}$ (3)

$=\dfrac{x}{y}$

My question is how is it possible to go from (2) to (3). Because in (2) there are three fractions and two equal signs, yet they are merged in (3). I don't understand how this can be done. Thank you.

That does look strange, but it works like this. (a+a+a)/(b+b+b)= 3*a/3*b = a/b

a/b = ca/cb = (a+ca)/(b+cb)

I'd never looked at it this way before.

 

[RChristenk]

I get what you are saying by itself, but how does the question follow this pattern? It makes no sense to set $a=\dfrac{2x-3y}{3z+y}$ for example. Thanks.

 

[PeroK]

Homework Statement: If $\dfrac{2x-3y}{3z+y}=\dfrac{z-y}{z-x}=\dfrac{x+3z}{2y-3x}$, prove that each of these ratios is equal to $\dfrac{x}{y}$
Relevant Equations: Basic fraction and algebra manipulation

I have the solution:

$\dfrac{2x-3y}{3z+y} = \dfrac{z-y}{z-x} = \dfrac {x+3z}{2y-3x}$ (1)

$\dfrac{2x-3y}{3z+y} = \dfrac{3(z-y)}{3(z-x)} = \dfrac {x+3z}{2y-3x}$ (2)

$= \dfrac{(2x-3y)-3(z-y)+x+3z}{(3z+y)-3(z-x)+(2y-3x)}$ (3)

$=\dfrac{x}{y}$

My question is how is it possible to go from (2) to (3). Because in (2) there are three fractions and two equal signs, yet they are merged in (3). I don't understand how this can be done. Thank you.

If we let the common ration be $r$, then we have something of form:
$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} =r$$Hence:
$$a_1+a_2 + a_3 = r(b_1+b_2+b_3)$$

 

[RChristenk]

But why would it be a negative sign in front of $-3(z-y)$? Shouldn't it be $= \dfrac{(2x-3y)+3(z-y)+x+3z}{(3z+y)+3(z-x)+(2y-3x)}$? And also why multiply it by 3 when the original ratio was $\dfrac{z-y}{z-x}$. Thanks.

 

[PeroK]

But why would it be a negative sign in front of $-3(z-y)$? Shouldn't it be $= \dfrac{(2x-3y)+3(z-y)+x+3z}{(3z+y)+3(z-x)+(2y-3x)}$? And also why multiply it by 3 when the original ratio was $\dfrac{z-y}{z-x}$. Thanks.

Why not multiply by $-3$? Why do any algebraic steps in a proof?

 

[Hornbein]

I also found this confusing and weird. I'm pretty sure there are no solutions to these equations. That is, there are no values of x,y,and z that make these equations true. It's a clever artificial example cooked up to make a point, which I by the way found enlightening.

It's based on the observation that ca/cb = a/b for all values of c except zero.

I can set a to be anything I like, within reason. It can be a polynomial if I feel like it. So

ca/cb = da/db = ea/eb for all b,c,d,e not equal to zero. Now in this case I don't even know what a or b are at all, I only propose that they "exist" in a sense. This is a pretty weird existence, as I believe there are no solutions to this system of equations, but that's all I need. Then I use (c+d+e)a/(c+d+e)b = a/b, do some algebraic canceling of terms, and lo and behold I've found the value of a/b, which is x/y.

 

[Hornbein]

It's also worth noting that the problem is not to prove the three ratios equivalent. This is given, assumed. The goal is to prove that it follows from this assumption that each is formally equivalent to x/y.

 

[Steve4Physics]

Hi @RChristenk. If you have not already sorted this out, maybe this will help…

The jump from equation (2) to (3) in the Post #1 solution combines several steps which should (IMO) have been made more explicit. So, expanding on what @PeroK has already said in post #4, here’s a somewhat detailed breakdown of what is going on...

We are given$$\frac {a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2} = R$$where $a_1 = 2x-3y, ~~a_2 = 3z +y,~b_1 = z-y,~~b_2=z-x,~c_1=x+3z$$c_2 = 2y- 3x$. We wish to show that $R = \frac xy$.

$a_1 = Ra_2,~~~~b_1 = Rb_2,~~~~c_1 = Rc_2$

Add these 3 equations to give:
$a_1+ b_1 + c_1 = R(a_2 + b_2 + c_2)$

Then $R = \frac {a_1+ b_1+c_1}{a_2~b_2+ c_2}$

Now the ‘tricky’ bit; this is where the multiplication '$-3$' comes in. We note that
$a_1+b_1+c_1=(2x-3y)+(z-y)+(x+3z)=3x-4y+4z$ and
$a_2+b_2+c_2=(3z+y)+(z- x)+(2y-3x)=-4x+3y+4z$

This doesn’t look promising but we know that we are aiming to show that $R=\frac xy$ which is independent of $z$. With some inspection and creative thinking, we spot (see below) that we can get rid of the ‘$z$’s if we let $b_1’ = -3b_1$ and $b_2’ = -3b_2$ and consider an equivalent initial set of equations:

$\frac {a_1}{a_2} = \frac {b_1’}{b_2’} = \frac {c_1}{c_2} = R$

This gives $R = \frac {a_1+ b_1'+c_1}{a_2~b_2'+ c_2}$

We now get
$a_1+ b_1’ + c_1 = 2x-3y+(-3)(z-y)+x +3z = 3x$
$a_2 + b_2’ + c_2 = 3z+y+(-3)(z-x) + 2y-3x = 3y$

By multiplying the $b$-terms by -3 we make the $z$'s cancel-out. And as a (predictable) bonus we are left with only $x$ in the numerator and $y$ in the denominator.

$R = \frac {3x}{3y} = \frac xy$.

Sorry if that's rather long-winded.

 

[RChristenk]

Hi @RChristenk. If you have not already sorted this out, maybe this will help…

The jump from equation (2) to (3) in the Post #1 solution combines several steps which should (IMO) have been made more explicit. So, expanding on what @PeroK has already said in post #4, here’s a somewhat detailed breakdown of what is going on...

We are given$$\frac {a_1}{a_2} = \frac {b_1}{b_2} = \frac {c_1}{c_2} = R$$where $a_1 = 2x-3y, ~~a_2 = 3z +y,~b_1 = z-y,~~b_2=z-x,~c_1=x+3z$$c_2 = 2y- 3x$. We wish to show that $R = \frac xy$.

$a_1 = Ra_2,~~~~b_1 = Rb_2,~~~~c_1 = Rc_2$

Add these 3 equations to give:
$a_1+ b_1 + c_1 = R(a_2 + b_2 + c_2)$

Then $R = \frac {a_1+ b_1+c_1}{a_2~b_2+ c_2}$

Now the ‘tricky’ bit; this is where the multiplication '$-3$' comes in. We note that
$a_1+b_1+c_1=(2x-3y)+(z-y)+(x+3z)=3x-4y+4z$ and
$a_2+b_2+c_2=(3z+y)+(z- x)+(2y-3x)=-4x+3y+4z$

This doesn’t look promising but we know that we are aiming to show that $R=\frac xy$ which is independent of $z$. With some inspection and creative thinking, we spot (see below) that we can get rid of the ‘$z$’s if we let $b_1’ = -3b_1$ and $b_2’ = -3b_2$ and consider an equivalent initial set of equations:

$\frac {a_1}{a_2} = \frac {b_1’}{b_2’} = \frac {c_1}{c_2} = R$

This gives $R = \frac {a_1+ b_1'+c_1}{a_2~b_2'+ c_2}$

We now get
$a_1+ b_1’ + c_1 = 2x-3y+(-3)(z-y)+x +3z = 3x$
$a_2 + b_2’ + c_2 = 3z+y+(-3)(z-x) + 2y-3x = 3y$

By multiplying the $b$-terms by -3 we make the $z$'s cancel-out. And as a (predictable) bonus we are left with only $x$ in the numerator and $y$ in the denominator.

$R = \frac {3x}{3y} = \frac xy$.

Sorry if that's rather long-winded.

Thanks a lot Steve! Now I finally understand what's happening here!

 

[PeroK]

I also found this confusing and weird. I'm pretty sure there are no solutions to these equations. That is, there are no values of x,y,and z that make these equations true.

$x=y=1$, $z=-\frac 2 3$

 

[PeroK]

PS only solutions are $x = y, z =-\frac 2 3 x$. For any $x \ne 0$.