6<\(\frac{4-x}{5}\)
\(\frac{26+x}{5}\) <0
x<-26
let see for the given value of x these options are always true
I x<26
values given x<-26
for every value of x
x is always <26 (suppose x=-27,-28,-29..........)
all these values are always less than 26
so this option must be true always
II. |x+19|>7
values given x<-26
for every value of x (suppose -27,-28,-29.......)
|x+19| =|-27+19| = 8, |-28+19| = 9, |-29+19| = 10 all values are greater than 7
so for every value of x this option must be
...
\(\frac{26+x}{5}\) <0
x<-26
let see for the given value of x these options are always true
I x<26
values given x<-26
for every value of x
x is always <26 (suppose x=-27,-28,-29..........)
all these values are always less than 26
so this option must be true always
II. |x+19|>7
values given x<-26
for every value of x (suppose -27,-28,-29.......)
|x+19| =|-27+19| = 8, |-28+19| = 9, |-29+19| = 10 all values are greater than 7
so for every value of x this option must be
...










