How do you solve #\frac { 3} { 8} - \frac { a + 3} { 4} \leq \frac { a - 1} { 2}#?

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Answer 1 · 2017-01-03T17:24:03

#a>= 1/6#

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#3/8 <= (a-1)/2 +(a+3)/4#

Or, #3/8<=(2a-2+a+3)/4#

Or, #3/8<= (3a+1)/4#. Now multiply the inequality by 8

#3<= 6a+2#

#1<=6a#

Therefore #a>= 1/6#