What is 5/8x<10?

I would like to have this done by today or so. Thank you!

4 Answers
Feb 22, 2018

The answer is #x<16#.

Explanation:

You can isolate #x# by performing the same actions on both sides of the inequality. In this case, first, multiply by #8#, then divide by #5#.

#5/8x<10#

#5/8xcolor(blue)(*8)<10color(blue)(*8)#

#5/color(red)cancel(color(black)8)x*color(red)cancel(color(black)8)<10*8#

#5x<10*8#

#5x<80#

#(color(red)cancel(color(black)5)x)/color(red)cancel(color(black)5)<80/5#

#x<80/5#

#x<16#

This is the solution to the inequality.

Feb 22, 2018

#x>1/16#

Explanation:

When working with these inequalities, we can treat them as regular algebraic equations when performing operations and moving the variable; however, we need to remember the less than sign:

#5/(8x)<10#

#5<(10)(8x)# (Multiply both sides by #8x#, this causes #8x# to cancel out on the left side)

#5<80x#

#1<(80/5)x# (Divide both sides by #5#, this causes #5# to cancel out on the left side)

#1<16x# (Simplify)

# 1/16<x # (Divide both sides by #16#. This causes #16# to cancel out on the right side)

If #1/16# is less than #x#, #x# must be greater than #1/16.# This is why we must remember the sign.
#x>1/16#

Feb 22, 2018

#x < 16#

Explanation:

As opposed to the algebraic equations you may be used to dealing with, an inequality instead describes a range of possible values for #x#.

If #x < 3#, #x# could be #2#, or #-1#, or even #-pi# because all of these values for #x# satisfy the condition that #x# must be less than #3#.

#color(red)(5/8)x < 10#

Much like an equation, you can perform identical actions to both sides of the inequality. The sole difference is that when multiplying or dividing by a negative number, the sign swaps. This is best illustrated with an example.

#3 < 5#, but #-3 > - 5#

Essentially, multiplying or dividing by a negative number flips the number line, in a sense.

#x < 10 * color(red)(8/5)#

#x < 16#

Feb 22, 2018

See below for an explanation:

Explanation:

The equation is #5/8x<10#

Multiply #8/5# to both sides:

#8/5*5/8x<10*8/5=>#

#cancel(8/5*5/8)x<16#

#x<16#

The reason why I multiplied #8/5# is that when multiplying a number with its reciprocal, it gives you #1#, enabling you to get the answer.