Question 146472
if {{{(bx - ay)/b = (cy - bz)/c = (az - cx)/a}}} and given that {{{bx!=ay}}}
{{{cy!=bz}}} and {{{az!=cx}}}, which of the follwing is true...??

(1){{{a+b=c}}}   (2){{{a+b+c=0}}}   (3){{{ab+bc+ca=0}}}    (4){{{ab+bc-ca=0}}}
<pre><font size = 4 color = "indigo"><b>
None of those are true necessarily.

Here is a counter-example which shows that:

{{{a=x=2}}}, {{{b=y=1}}}, {{{c=z=1}}}

We show that in this case:

{{{(bx - ay)/b = (cy - bz)/c = (az - cx)/a}}}

is true by substituting:

{{{(1*2 - 2*1)/1 = (1*1 - 1*1)/1 = (2*1 - 1*2)/2}}}

{{{0=0=0}}}

Now we show that

{{{bx!=ay}}} is true by substituting:

{{{1*2!=2*1!}}}

{{{2=2}}}

Then we show that

{{{cy!=bz}}} is true by substituting:

{{{1*1!=1*1!}}}

{{{1=1}}}

And finally we show that

{{{az!=cx}}} is true by substituting:

{{{2*1!=1*2!}}}

{{{2=2}}}

Now look at the four choices:

choice (1)  {{{a+b=c}}}   

Substituting:  {{{2+1=1}}} which is false

choice (2)  {{{a+b+c=0}}} 

Substituting:  {{{2+1+1=0}}} which is false

choice (3)  {{{ab+bc+ca=0}}}  

Substituting: {{{2*1+1*1+1*2=0}}} which is false

choice (4){{{ab+bc-ca=0}}}

Substituting: {{{2*1+1*1-1*2=0}}} which is also false.

So you can tell your teacher that this is a bogus problem.

Edwin</pre>