Question 331485
<pre><b>
You don't need to use the quadratic formula,
as the other tutor used, because it leads to 
a factorable quadratic:

{{{system(xy=1,
2x-y=1)}}}

Solve the second equation for y:

{{{matrix(3,1,
   2x-y=1,
   2x-1=y,
   y=2x-1
)}}} 

Substitute {{{(2x-1)}}} in the first equation in the system

{{{matrix(8,5,
xy=1," "," ", " "," ",
x(2x-1)=1," "," ", " "," ",
2x^2-x=1," "," ", " "," ",
2x^2-x-1=0," "," ", " "," ",
(2x+1)(x-1)=0," "," ", " "," ",
2x+1=0," "," ", " ",x-1=0,
2x=-1," "," ", " ",x=1,
x=-1/2," "," ", " "," ")}}}

So there are two solutions for x.  We
must find a value of y for each of them,
by substituting each in {{{y=2x-1}}}

Substituting {{{x=-1/2}}},
{{{y=2x-1}}}
{{{y=2(-1/2)-1}}}
{{{y=-1-1}}}
{{{y=-2}}}

So one solution is (x,y) = ({{{-1/2}}},-2) 

Substituting {{{x=1}}},
{{{y=2x-1}}}
{{{y=2(1)-1}}}
{{{y=2-1}}}
{{{y=1}}}

So the other solution is (x,y) = (1,1)

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

Checking (x,y) = ({{{-1/2}}},-2)

{{{system(xy=1,
2x-y=1)}}}

{{{system((-1/2)(-2)=1,
2(-1/2)-(-2)=1)}}}

{{{system(1=1,
-1+2=1)}}}

{{{system(1=1,
1=1)}}} 

Checking (x,y) = (1,1)

{{{system(xy=1,
2x-y=1)}}}

{{{system((1)(1)=1,
2(1)-(1)=1)}}}

{{{system(1=1,
2-1=1)}}}

{{{system(1=1,
1=1)}}}

Edwin</pre>