The sum of three numbers is 98 . The third number is 8 less than the first. The second number is 3 times the third. What are the numbers?

Let the three numbers be denoted as #n_1#, #n_2#, and #n_3#.

"The sum of three numbers is #98#"

#[1] => n_1+n_2+n_3 = 98#

"The third number is #8# less than the first"

#[2] =>n_3 = n_1 - 8#

"The second number is #3# times the third"

#[3] =>n_2 = 3n_3#

We have #3# equations and #3# unknowns, so this system may have a solution that we can solve for. Let's solve it.

First, let's substitute #[2] -> [3]#

#n_2 = 3(n_1 - 8)#

#[4]=> n_2 = 3n_1 - 24#

We can now use #[4]# and #[2]# in #[1]# to find #n_1#

#n_1 + (3n_1-24) + (n_1-8) = 98#

#n_1 + 3n_1 - 24 + n_1 - 8 = 98#

#5n_1 -32 = 98#

#5n_1 = 130#

#[5]=>n_1 =26#

We can use #[5]# in #[2]# to find #n_3#

#n_3 = 26 - 8 #

#[6]=>n_3 = 18#

Lastly, we can use #[6]# in #[3]# to find #n_2#

#n_2 = 3(18)#

#[7]=>n_2 = 54#

Hence, our solution from #[5], [6], [7]# is:

#n_1 = 26#

#n_2 = 54#

#n_3 = 18#

let a= first no.;b= second no. and c= third no.

now given(The third number is 8 less than the first. )
then,#c=a-8#
also given(The second number is 3 times the third)
then,#b=3*c ;=>3*(a-8); =>3*a-24#
now adding
#a+b+c=98# (given)
putting the values of b and c

#=>a+3*a-24+a-8=98#
#=>5*a-32=98#
#=>5a=130;=>a=26#

now #c=a-8; =>26-8 ; =>18#
and #b=3*c ; =>3*18; =>54#