Question 249402
Do you mean: {{{(1/2)y+(1/4)y>=-6}}} ...OR...  {{{(1/(2y))+(1/(4y))>=-6}}}  ??


***  Please note: When typing problems in computers, please use parenthesis to distinguish numerators from denominators ...and... fractions from variables that are not inside the fractions.  This way there are no misunderstandings.  ***


Solving for: {{{(1/2)y+(1/4)y>=-6}}}
Multiply by 4 on both sides of the inequality to get rid of fractions.
{{{4((1/2)y+(1/4)y)>=4*(-6)}}}
We get:
{{{2y+y>=-24}}}
Simplify:
{{{3y>=-24}}}
Divide by 3 on both sides to determine the values of y.
{{{y>=-8}}}


Solving for:  {{{(1/(2y))+(1/(4y))>=-6}}}
Multiply by 4y on both sides of inequality to get rid of fractions.
{{{4y((1/(2y))+(1/(4y)))>=4y(-6)}}}
We get:
{{{2y+1>=-24y}}}
Combine like terms:
{{{2y+24y>=-1}}}
Simplify:
{{{26y>=-1}}}
Divide by 26 on both sides to determine the values of y.
{{{y>=-(1/26)}}}