# 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?

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 $98$"

$\left[1\right] \implies {n}_{1} + {n}_{2} + {n}_{3} = 98$

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

$\left[2\right] \implies {n}_{3} = {n}_{1} - 8$

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

$\left[3\right] \implies {n}_{2} = 3 {n}_{3}$

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 $\left[2\right] \to \left[3\right]$

${n}_{2} = 3 \left({n}_{1} - 8\right)$

$\left[4\right] \implies {n}_{2} = 3 {n}_{1} - 24$

We can now use $\left[4\right]$ and $\left[2\right]$ in $\left[1\right]$ to find ${n}_{1}$

${n}_{1} + \left(3 {n}_{1} - 24\right) + \left({n}_{1} - 8\right) = 98$

${n}_{1} + 3 {n}_{1} - 24 + {n}_{1} - 8 = 98$

$5 {n}_{1} - 32 = 98$

$5 {n}_{1} = 130$

$\left[5\right] \implies {n}_{1} = 26$

We can use $\left[5\right]$ in $\left[2\right]$ to find ${n}_{3}$

${n}_{3} = 26 - 8$

$\left[6\right] \implies {n}_{3} = 18$

Lastly, we can use $\left[6\right]$ in $\left[3\right]$ to find ${n}_{2}$

${n}_{2} = 3 \left(18\right)$

$\left[7\right] \implies {n}_{2} = 54$

Hence, our solution from $\left[5\right] , \left[6\right] , \left[7\right]$ 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
$a + b + c = 98$ (given)
$\implies a + 3 \cdot a - 24 + a - 8 = 98$
$\implies 5 \cdot a - 32 = 98$
=>5a=130;=>a=26
now c=a-8; =>26-8 ; =>18
and b=3*c ; =>3*18; =>54