Question 893536
Gather like terms,
{{{(5/4)a^2-3a-1/2+(3/8)a^2+(1/2)a+2=(5/4+3/8)a^2+(-3+1/2)a+(-1/2+2)}}}
Now get a common denominator for each group of sums,
{{{(5/4)a^2-3a-1/2+(3/8)a^2+(1/2)a+2=(10/8+3/8)a^2+(-6/2+1/2)a+(-1/2+4/2)}}}
{{{(5/4)a^2-3a-1/2+(3/8)a^2+(1/2)a+2=(13/8)a^2+(-5/2)a+(3/2)}}}
{{{(5/4)a^2-3a-1/2+(3/8)a^2+(1/2)a+2=highlight((13/8)a^2-(5/2)a+3/2)}}}
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{{{(1/2)x^2y+(1/3)xy-1+x^2y-(2/3)xy+5=(1/2+1)x^2y+(1/3-2/3)xy+(-1+5)}}}
{{{(1/2)x^2y+(1/3)xy-1+x^2y-(2/3)xy+5=(1/2+2/2)x^2y+(1/3-2/3)xy+(-1+5)}}}
{{{(1/2)x^2y+(1/3)xy-1+x^2y-(2/3)xy+5=(3/2)x^2y+(-1/3)xy+(4)}}}
{{{(1/2)x^2y+(1/3)xy-1+x^2y-(2/3)xy+5=highlight((3/2)x^2y-(1/3)xy+4)}}}