Question 1186839


{{{(y-3)/8>=(y+2)/6 }}}....just like you would do with equality, cross multiply

{{{6(y-3)>=8(y+2) }}}

{{{6y-18>=8y+16 }}}

{{{-16-18>=8y-6y}}}

{{{-34>=2y}}}

{{{-34/2>=y}}}

{{{-17>=y}}}

or

{{{y<=-17}}}

Interval notation:

({{{-infinity}}},{{{ -17}}}]

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