Question 701519
It's not perfectly clear to me, but
I assume that the problem is {{{system((1/2)x + (2/3)y = 1,(1/4)x - (1/3)y = 5/2)}}}, and I'll solve that.
 
If I had a choice, I would not solve that system by substitution.
{{{(1/4)x - (1/3)y = 5/2 }}} --> {{{(1/4)x = (1/3)y+5/2 }}} --> {{{x=(4/3)y+10}}}
Substituting {{{(4/3)y+10}}} for {{{x}}} in {{{(1/2)x + (2/3)y = 1}}} we get
{{{(1/2)((4/3)y+10) + (2/3)y = 1}}} --> {{{(2/3)y+5+ (2/3)y = 1}}} --> {{{(4/3)y=1-5}}} --> {{{(4/3)y=-4}}} --> {{{(3/4)(4/3)y=-4(3/4)}}} --> {{{highlight(y=-3)}}}.
Then, substituting {{{(-3)}}} for {{{y}}} in {{{x=(4/3)y+10}}}, we get
{{{x=(4/3)(-3)+10}}} --> {{{x=-4+10}}} --> {{{highlight(x=6)}}}.
 
If I could choose how to solve it,
I would first transform the equations,
so as to get equivalent equations without fractions.
{{{(1/2)x + (2/3)y = 1}}} --> {{{3x+4y=6}}} (multiplying everything times 6)
{{{(1/4)x - (1/3)y = 5/2 }}} --> {{{3x-4y=30}}} (multiplying everything times 12)
Then I would add both equations to get
{{{6x=36}}} --> {{{highlight(x=6)}}}.
I would also subtract the second equation from the first one to get
{{{8y=-24}}} --> {{{highlight(y=-3)}}}.
 
IF THE PROBLEM WAS
{{{system(1/2x + 2/3y = 1,1/4x - 1/3y = 5/2 )}}},
That's the same as
{{{system((1/2)(1/x) + (2/3)(1/y) = 1,(1/4)(1/x) - (1/3)(1/y) = 5/2 )}}}.
That could be solve the same way for {{{1/x}}} and {{{1/y}}} to get
{{{1/x=6}}} and {{{1/y=-3}}}
Then the solutions would be {{{x=1/6}}} and {{{y=-1/3}}}.