Question 837118
Clarifying what you tried to do:

{{{X<>x}}}.
Choose which way you are using in any given problem and use the variable consistently.  
{{{x=x}}} and {{{X=X}}}, but {{{x<>X}}} and {{{X<>x}}}.


Your given system:
{{{x=y+8}}} and {{{5x+3y=12}}};
Most of what you show looks good.
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You have a formula for x as one of the given equations.
Substitute...
{{{5(y+8)+3y=12}}}
{{{5y+40+3y=12}}}
{{{8y+40=12}}}
{{{8y=12-40}}}
{{{8y=-28}}}
{{{2y=-7}}}
{{{highlight(y=-7/2)}}} just as you found.  (You showed one arithmetic mistake, like you intended division by 4 but wrote division by 8).
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Use the value for y to find the value of x, from your original first equation:
{{{x=y+8}}}
{{{x=-(7/2)+8}}}
{{{x=-(7/2)+8(2/2)}}}
{{{x=-(7/2)+16/2}}}
{{{highlight(x=9/2)}}}
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SUMMARY OF ANSWER RESULTS: {{{x=9/2}}} and {{{y=-(7/2)}}}.