How do you solve #12x – 5x < 6(x + 7)#?

2 Answers
May 3, 2018

Answer:

#x < 42#

Explanation:

#"simplify left side and distribute parenthesis"#

#rArr7x<6x+42larrcolor(blue)"subtract 6x from both sides"#

#rArrx<42" is the solution"#

#x in(-oo,42)larrcolor(blue)"in interval notation"#

May 3, 2018

Answer:

You get #x<42#

Explanation:

I would simplify the left side, since #12-5x = 7x#.

Therefore #7x<6(x+7)#
Now #6(x+7)# is the same as multiplying each term in the parenthesis with the number outside, i.e.
#6(x+7) = 6x + 6*7 = 6x+42#
Therefore we have
#7x<6x+42#

So 7 of something should be smaller than 6 of something pluss 42.

I hope it then is easy to see that this is fulfilled if 1 of that something is smaller than 42,
i.e.
#x<42#

Normally we would do it this way, deducting the same amount from each side:
#7x-6x < 6x+42-6x#
Which would give
#x<42#