How do you solve and write the following in interval notation: #2/5x<6# and #-1/2x <= -10#?

1 Answer
Jul 15, 2018

Answer:

#x in(-oo,15)uu[20, oo)#

# x in RR-[15,20)#

Explanation:

We know that,

#color(red)((I)"if a < b and c > 0# #" then "# #color(red)(ac < bc and "a/c < b/c #

#color(blue)((II)"if a <= b and c < 0# #" then"# #color(blue)( ac >= bc and "a/c >= b/c #

Here ,

#2/5x < 6 and -1/2x <= -10#

#(i) 2/5x < 6#
Multiplying both side by #color(red)(5/2#

#color(red)(5/2)2/5x < color(red)( 5/2)6color(red)( to Apply(I)#

#=>x < 15#

#=>x in (-oo, 15)#

#(ii)-1/2x <= -10#

Multiplying both sides by #color(blue)((-2)#

#color(blue)((-2))(-1/2x) >=color(blue)((-2))(-10)color(blue)(toApply(II)#

#=>x >= 20#

#=>x in[20 , oo)#

From #(i) and (ii) #we get

#x in (-oo,15) or x in [20, oo)#

#=>x in(-oo,15)uu[20, oo)#

#=> x in RR-[15,20)#