Question 835281
This could be easier if use substitution.  Maybe first multiply the first equation by 2, and the second equation by 15... you still might be able to use Elimination.


{{{(5c/4)2+2*d=11}}}
{{{5c/2+2d=11}}}
{{{(5/2)c+2d=11}}}
One more step, multiply by 2 b.s.
{{{5c+4d=22}}}
-
{{{15c+(d/3)15=14}}}
{{{15c+5d=14}}}


I really am more interested in using Elimination Method.  The system is now displayable as:
{{{5c+4d=22}}} and {{{15c+5d=14}}}.
Multiplying the first of these by 3 on both sides, our system is:
{{{15c+12d=66}}} and {{{15c+5d=14}}}
READY TO ELIMINATE c.


Continuing with Elimination Method,
{{{(15c+12d)-(15c+5d)=66-14}}}
{{{15c-15c+12d-5d=52}}}
{{{7d=52}}}
{{{highlight(d=52/7)}}} or {{{highlight(d=7&3/7)}}}
-
Return to almost any equation you want, find c and plug in the value for d.
c  = 14/15-d/3
c=14/15-(52/7)/3
c=14/15-52/21, lcd is 3*5*7
c=(14/15)(7/7) - (52/21)(5/5)
c=(98-260)/105
c=-162/105=-54/35
{{{highlight(c=-(54/35)=-1&19/35)}}}