Question 174787
For the system of equations
{{{system(ax + by = c,
dx + ey = f)}}}
Find the conditions on a,b,c,d,e and f, where 
b and e are nonzero, such that the equations have 
<pre><font size = 4 color = "indigo"><b>
There are more than one way to do this, and
the method used depends on what course you
are taking.  It could be done with determinants,
matrices, or neither.  And there are different 
ways in each of those.  I will arbitrarily do it 
one way with neither.  If you're studying another 
method then post again telling us what method 
you are studying. 

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Equations like these

i.  {{{0x=0}}} and {{{0y=0}}} 

have infinitely many solutions
since any number may be 
substituted for x or y and the
the equation will 
always become {{{0=0}}},
which will always be true.

Equations like these

ii. {{{0x=r}}} and {{{0y=s}}}

where neither r nor s is 0

cannot have any solutions,
because the left side will
always be 0 and the right
side will be something other
than 0, which will never be 
true.

iii.  Equations like these

ii. {{{px=r}}} and {{{qy=s}}}

where neither p nor q is 0  will
have exactly one unique solution, 
for since neither the coefficient
of x and y is  0, then we can
divide both sides by that non-zero
coefficient and get one and only
one solution.

----------------------

Let's start to solve the system by
elimination:

{{{system(ax + by = c,
dx + ey = f)}}}

Multiply the first equation through by {{{-d}}}
and multiply the second equation through by {{{""+a}}}
to eliminate the x's

{{{system(-adx - bdy = -cd,
""+adx + aey = af)}}}

Adding corresponding terms, the x-terms cancel
out and we have:

    {{{aey-bdy = af-cd}}}

Factor out y on the left:

    {{{y(ae-bd) = af-cd}}}

------

Now we eliminate the y-terms.

{{{system(ax + by = c,
dx + ey = f)}}}

Multiply the first equation through by {{{""+e}}}
and multiply the second equation through by {{{-b}}}
to eliminate the x's

{{{system(""+aex + bey = ce,
-bex - bey = -bf)}}}

Adding corresponding terms, the y-terms cancel
out and we have:

    {{{aex-bex = ce-bf}}}

Factor out x on the left:

    {{{x(ae-bd) = ce-bf}}}

So we have the system:

{{{system(x(ae-bd) = ce-bf, y(ae-bd)=af-cd)}}}
</pre></font></b>
i.   Infinitely many solutions
<pre><font size = 4 color = "indigo"><b>

Here we need to have a case of

{{{system(0x = 0, 0y=0)}}}, so we set the 
coefficient of x and y on the left, which are
both {{{(ae-bd)}}}, equal to 0:

{{{ae-bd=0}}} or {{{ae=bd}}}

We also set each right side equal to 0.
Setting the first right side = 0,

{{{ce-bf=0}}} or {{{ce=bf}}} 

Setting the second right side = 0,

{{{af-cd=0}}} or {{{af=cd}}}

Putting these results toghether, these fractions
are all equal, so the answer to (i) is

{{{system(ae=bd, ce=bf,af=cd)}}}
</pre></font></b>
ii. no solution
<pre><font size = 4 color = "indigo"><b>
{{{0x=p}}} and {{{0y=q}}}

where neither p nor q is 0, so

{{{system(x(ae-bd) = ce-bf, y(ae-bd)=af-cd)}}}

This system is like (i) in that we have

{{{ae-bd=0}}} or {{{ae=bd}}}

However it is unlike (i) in that we must 
require that the right sides NOT equal 0,

Setting the first right side NOT equal 0,

{{{ce-bf<>0}}} or {{{ce<>bf}}} 

Setting the second right side NOT equal 0,

{{{af-cd<>0}}} or {{{af<>cd}}}

Putting these results together, the answer 
to (ii) is

{{{system(ae=bd, ce<>bf,af<>cd)}}}

---
</pre></font></b>
iii. A unique solution
<pre><font size = 4 color = "indigo"><b>
There will be exactly one
unique solution if the coefficient
of x and y is not 0, for then we can
divide both sides by that non-zero
coefficient and get one and only
one solution.

{{{system(x(ae-bd) = ce-bf, y(ae-bd)=af-cd)}}}

So we require that {{{(ae-bd)<>0}}}, so that
we can divide both sides of both equations
by {{{ae-bd}}} and get just one value for x
and just one value for y.

So we just have to require that

{{{ae-bd<>0}}} or {{{ae<>bd}}}

So the answer to (iii) is

{{{ae<>bd}}} and there are no restrictions on
{{{c}}} or {{{f}}}.

Edwin</pre>