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?

2 Answers
Jun 25, 2018

n_1 = 26

n_2 = 54

n_3 = 18

Explanation:

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

Jun 25, 2018

first no. =26; second no.=18;third no.=54

Explanation:

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