How do you solve and write the following in interval notation: #3x+4≥5x-8#?

2 Answers
Jun 22, 2016

Answer:

#x in (-oo,6]#

Explanation:

Remember you can add/subtract the same amount and multiply/divide by any amount greater than zero to both sides of an inequality without invalidating the inequality.

#3x+4 >= 5x-8#
#color(white)("XXX")3x+12 >= 5x#

#color(white)("XXX")12 >= 2x#

#color(white)("XXX")6 >= x#

or (reversing so #x# is on the left, which is more common)
#color(white)("XXX")x <= 6#

Jun 22, 2016

Answer:

You may add or subtract on both sides of the inequality.

Explanation:

Add #8# to both sides:
#3x+4+8>=5x-cancel8+cancel8->3x+12>=5x##

Subtract #3x# from both sides:
#cancel(3x)-cancel(3x)+12>=5x-3x->12>=2x->#

Divide by #2#, a positive number:
#6>=x->x<=6#

#->x in [6,oo)#, where #[# means 'inclusive'.

Note: you may be used to a different notation, but I think the above is clear enough to translate.