find three consecutive integers such that the sum between the first integer and twice the second integer is equal to ten less than the third integer

Answer 1

Let x = the smallest of the three consecutive integers

Then the others are x + 1 and x + 2.

So, x + 2(x + 1) = (x + 2) - 10

3x + 2 = x - 8

2x = -10

x = -5

The integers are -5, -4, and -3.

Answer 2

Integers are X, Y, and Z

"consecutive" means: Y = X + 1, and Z = Y + 1

The other information given:

"the sum between the first integer and twice the second integer is equal to ten less than the third integer"

gives us this:

X + 2Y = Z - 10 (equation #3)

From the first pair of equations:

X = Y - 1

Z = Y + 1

Substitute these into equation #3:

(Y - 1) + 2Y = (Y + 1) - 10

Rearrange and simplify:

3Y - 1 = Y - 9

2Y = -8

Y = -4

Numbers are -5, -4, and -3

Answer 3

The 3 consecutive integers are

1st: n

2nd: n + 1

3rd: n + 2

Then

n + 2(n + 1) = (n + 2) - 10

Can you solve for n and finish from here?

Answer 4

a+2b=c-10

2+2*4=20-10

There's many more answers.