Math problem about three integers

Question

Find three consecutive integers such that the sum of the first and twice the second is 110 minus 3 times the third

Answer 1

The 3 consecutive integers are

n

n + 1

n + 2

The given information translates into

n + 2(n + 1) = 110 - 3(n + 2)

n + 2n + 2 = 110 - 3n - 6

3n + 2 = 104 - 3n

6n = 102

n = 102/6 = 17

n + 1 = 18

n + 2 = 19

Test the answer.

Answer 2

Hi Gabrielle,

At first, this may seem complicated because we have to translate the words to math. But it can be straight-forward! Let's do it together.

"Find three consecutive integers" does not mean you have 3 different variables. Because the integers (whole numbers) are consecutive (like 1, 2, 3), you can define the first integer as "x", the second integer as "x+1" and the third integer as "x+2."

"such that the sum of the first (integer) and twice the second (integer) is 110 minus 3 times the third (integer)"

Sum means we're adding; when you see "is" that is your equals sign.

So:

x + 2(x+1) = 110 - 3(x+2)

x + 2x + 2 = 110 -3x - 6

3x + 2 = 104 - 3x

3x + 3x = 104 - 2

6x = 102

x = 102/6

x = 17

We now know that the first integer is 17, and your final answer includes the next 2 consecutive integers.

Plug them back in to the equation to double-check your work.

Answer 3

The three integers are n, n+1, and n+2

n + 2(n+1) is "the sum of the first and twice the second"

110 - 3(n+2) is "110 minus 3 times the third"

The problem asks to find n such that the above two expressions are equal:

n + 2(n+1) = 110 - 3(n+2)

n + (2n + 2) = 110 - (3n + 6)

3n + 2 = 110 - 3n - 6

6n = 102

n = 102/6

n = 17 --> the numbers are 17,18, and 19

Check: 17+2(18) = 17+36 = 53

and 110-3(19) = 110-57 = 53