How do you solve the inequality #3(t - 3) + 1 ≥ 7# and #2(t + 1) + 3 ≤ 1#?

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Answer 1 · 2017-08-06T16:38:14

See a solution process below:

Solve First Inequality

Begin by solving the first inequality:
#color(red)(3)(t - 3) + 1 >= 7#

#(color(red)(3) xx t) - (color(red)(3) xx 3) + 1 >= 7#

#3t - 9 + 1 >= 7#

#3t - 8 >= 7#

#3t - 8 + color(red)(8) >= 7 + color(red)(8)#

#3t - 0 >= 15#

#3t >= 15#

#(3t)/color(red)(3) >= 15/color(red)(3)#

#(color(red)(cancel(color(black)(3)))t)/cancel(color(red)(3)) >= 5#

#t >= 5#

Solve Second Inequality

Next, we can solve the second inequality for #t#:

#color(red)(2)(t + 1) + 3 <= 1#

#(color(red)(2) xx t) + (color(red)(2) xx 1) + 3 <= 1#

#2t + 2 + 3 <= 1#

#2t + 5 <= 1#

#2t + 5 - color(red)(5) <= 1 - color(red)(5)#

#2t + 0 <= -4#

#2t <= -4#

#(2t)/color(red)(2) <= -4/color(red)(2)#

#(color(red)(cancel(color(black)(2)))t)/cancel(color(red)(2)) <= -2#

#t <= -2#

**The Solution Is:

#t < -2#; #t >= 5#

Or, in interval notation:

#(-oo, -2]; [5, +oo)#