Why can't we do crossing over in inequalities?

If we have the inequality #12/(x-4)<=4#, why can't we multiply both sides by #x-4#?

1 Answer
Aug 5, 2018

Because you might be mutliplying both sides by a negative quantity.

Explanation:

Given any inequality, some operations that you can perform and preserve the truth or falsity of the inequality are:

  • Add or subtract the same expression on both sides.
  • Multiply or divide both sides by a positive expression.
  • Multiply or divide both sides by a negative expression and reverse the inequality (i.e. #<# becomes #>#, #>=# becomes #<=#, etc.).
  • Apply the same strictly monotonically increasing function to both sides.
  • Apply the same strictly monotonically decreasing function to both sides and reverse the inequality.

Given:

#12/(x-4) <= 4#

we would like to get #(x-4)# out of the denominator, but multiplying both sides by #(x-4)# would result in the wrong inequality when #x < 4#.

One option is to note that #x = 4# is not in the solution space (since it results in division by #0#), then multiply both sides by #(x-4)^2# for the cases where #x != 4#, since when #x != 4# we have #(x-4)^2 > 0#.

We find:

#12(x-4) <= 4(x-4)^2 = 4(x^2-8x+16)#

Then we can divide both ends by #4# to get:

#3x-12 <= x^2-8x+16#

Then subtract #3x-12# from both sides to get:

#0 <= x^2-11x+28 = (x-4)(x-7)#

Note this inequality would be true for #x in (-oo, 4] uu [7, oo)#

but the point #x=4# is excluded.

So the solution of the original inequality is:

#x in (-oo, 4) uu [7, oo)#