How do you solve #12x – 5x < 6(x + 7)#?
#"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"#
I would simplify the left side, since
Therefore we have
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,
Normally we would do it this way, deducting the same amount from each side:
Which would give