Question

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?

Answer 1

`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`

Answer 2

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`