The sum of three integers is 92. the second number is three times the first number. the third number is ten less than twice the first number. find the integers.

Hello from MrBillDoesMath!

17, 51, 24

Discussion:

Suppose the three integers are "a", "b", and "c". Then

a + b + c = 92  (*)

We are told that   b = 3a and c =  2a - 10. Substituting these values in (*) gives

a + (3a) + (2a -10) = 92  =>

(a + 3a + 2a) - 10 = 92 =>

6a                 - 10  = 92.

Add 10 to each side

6a - 10 + 10 = 92 + 10 =>

6a = 102 =>

a = 17

Then b = 3a = 3(17) = 51, and c = 2(17) - 10 = 24

a + b + c = 17 + 51  + 24  = 92

Thank you,

MrB

The three integers are  17, 51, 24

Step-by-step explanation:

Let us take x, y and z are the three integers.

x + y + z = 92 (1)

The second number is three times the first number.

y = 3x

The third number is ten less that twice the first number.

z = 2x  - 10

Now plug in y = 3x and z = 2x - 10 in the equation (1)

x + (3x) + (2x -10) = 92

x + 3x + 2x - 10 = 92

6x - 10 = 92

Add 10 on both sides, we get

6x - 10 + 10 = 92 + 10

6x = 102

Dividing both sides by 6, we get

x = 17

So, x = 17

y = 3 × 17 = 51

and

z = 2x - 10 = 2×17 - 10 = 34 -10 = 24

i dont know what you want me to answer for this but im guessing 2: 11?