How do you solve #8(a-2)<=10(a+2)#?

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Answer 1 · 2018-04-16T01:03:58

#a >= -18#

We solve inequalities similarly to equations.

#8(a-2) <= 10(a+2)#

First, let's distribute:
#8a - 16 <= 10a + 20#

Now subtract #8a# from both sides of the inequality:
#8a - 16 quadcolor(red)(-quad8a) <= 10a + 20 quadcolor(red)(-quad8a)#

#-16 <= 2a + 20#

Subtract #20# from both sides of the inequality:
#-16 quadcolor(red)(-quad20) <= 2a + 20 quadcolor(red)(-quad20)#

#-36 <= 2a#

Divide both sides by #2#:
#(-36)/color(red)(2) <= (2a)/color(red)(2)#

#-18 <= a#

Put #a# on the left side of the inequality:
#a >= -18#

This means that #a# must be more than or equal to #-18#.

Hope this helps!