How do you solve and write the following in interval notation: #(x-5)/3 - (4x+3)/6> -3#?

1 Answer
Jul 22, 2016

Answer:

#x in (-oo, 5/2)#

Explanation:

The first thing to do here is get rid of the denominators by multiplying the first fraction by #1 = 2/2# and #-3#, which can be thought of as a fraction with the denominator equal to #1#, by #12 = 6/6#.

The inequality becomes

#(x-5)/3 * 2/2 - (4x+3)/6 > -3 * 6/6#

#(2(x-5))/6 - (4x+3)/6 > -18/6#

This will be equivalent to

#2(x-5) - 4x - 3 > -18#

Expand the parentheses and group like terms to find

#2x - 10 - 4x - 3 > -18#

#-2x - 13 > -18#

#-2x > -5#

Now, you must divide both sides of the inequality by #-2# to isolate #x#. Do not forget that multiplying or dividing by a negative number changes the sign of the inequality!

In this case, you have

#> -> <#

and

#(color(red)(cancel(color(black)(-2)))x)/(color(red)(cancel(color(black)(-2)))) < (-5)/(-2)#

#x < 5/2#

This tells you that any value of #x in RR# that is smaller than #5/2# will be a solution to the given inequality. Mind you, #5/2# is not part of the solution interval.

In interval notation, this can be written as

#color(green)(|bar(ul(color(white)(a/a)color(black)(x in (-oo, 5/2))color(white)(a/a)|)))#

graph{x<5/2 [-10, 10, -5, 5]}