The sum of three numbers is 81. the second is twice the first and the third is 6 more than the second. find the numbers with workings?

VTPETERS2002

5 years ago

Let the three numbers be a, b, c, respectively.

The sum of the three numbers is 81 → a+b+c=81

The second number, b , is twice the first number, a → b=2a .

The third number, c , is 6 more than the second number, b → c=b+6 .

Let’s get this all into one variable. We can also re-write the value of c in terms of a . Since b=2a , we can re-write c=b+6 as c=2a+6 .

So now we re-write a+b+c=81 as a+(2a)+(2a+6)=81 , by replacing b with 2a and replacing c with (2a+6) .

Now we can simplify the equation a+(2a)+(2a+6)=81 as 5a+6=81 by combining like terms.

Then solve for a in 5a+6=81 .

5a=81−6

5a=75

a=75/5

a=15

Now plug in a=15 into b=2a to find the value of b .

b=2(15)

b=30

Now plug in b=30 into c=b+6 to find the value of c .

c=(30)+6

c=36

Hence the three numbers are 15, 30, 36 , respectively.