Question 938455
<pre>
We want to eliminate a letter.

5r+2t = 5
3r-7t = 3

Let's choose r to eliminate.

The coefficient of r in the first equation is 5.
The coefficient of r in the second equation is 3.
The least common multiple of 5 and 3 is 15.  
We want the terms in r to be equal in magnitude
but opposite in sign.  To do that, we multiply both
sides of the first equiation by -3 and both sides
of the second equation by 5.  We get

-3(5r+2t) = -3(5)
 5(3r-7t) =  5(3)

 -15r- 6t = -15
  15r-35t =  15

Next we add them vertically, term by term:

 -15r- 6t = -15
  15r-35t =  15
--------------
   0r-41t =   0
     -41t =   0
    
Divide both sides by -41

 {{{(-41t)/(-41)}}} = {{{0/(-41)}}}

        t = 0

Substitute 0 for t in either one of the 

    5r+2t = 5  
  5r+2(0) = 5
     5r+0 = 5
       5r = 5

Divide both sides by 5

    {{{5r/5}}} = {{{5/5}}}
        
        r = 1

Solution (r,t) = (1,0)

Edwin</pre>