Question #dbf22

1 Answer
Nov 27, 2015

Answer:

#x>8/3#

Explanation:

If you look at https://socratic.org/help/symbols you can find tips on formatting. Use the hash at both the start and end of any maths equations etc.

Given: #1/4(x+4)<1/5(2x+3)#

Use standard manipulation as if the < behaved the same way (sort of ! ) as an =. You have to be careful about what you are doing if you have any multiplications involving negatives. You don't!

#=> 5(x+4)<4(2x+3)#

#5x+20<8x+12#

Collecting like terms

#20-12<8x-5x#

#8<3x#

Divide by 3

#8/3 < x#

Turn it round so that it has the variable on the left. Notice that you still have to make #x# bigger than #8/3# so that turns round as well.

#x>8/3#