Find 3 consecutive integers that the sum of the first and twice the third equal 2 more than 3 times the second

Answer 1

Just set up the algebra:

x = "the first number"
x + 1 = "the second number"
x + 2 = "the third number"

So, we can make the following equation to solve for the first number:

x + 2(x + 2) = 3(x + 1) + 2
==> distribute
x + (2x + 4) = (3x + 3) + 2
==> combine like terms
3x + 4 = 3x + 5
==> subtract 3x from both sides
0 ≠ 1
==> THERE IS NO SOLUTION TO THIS EQUATION

This is undefined. It appears that there is no solution to the equation you entered. Are you sure you typed it in correctly?

Answer 2

"3 consecutive integers"

x, x + 1, x + 2

"The sum of the first and twice the third"

x + 2(x + 2)

"is"

=

"2 more than 3 times the second"
3(x + 1) + 2

Put it all together, and you get

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

Expand, isolate x, solve.

x + 2x + 4 = 3x + 3 + 2
3x + 4 = 3x + 5
4 = 5, which is a false statement.

Therefore, there exists no integer (which is an answer I didn't expect). I either made an error in one step, or there really is no solution.

Answer 3

x
x+1
x+2

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

x + 2x +4 = 3x +3 +2

3x +4 = 3x + 5

no solution

Answer 4

let n-1, n, n+1 be the 3 integers.

(n-1) + 2(n+1) = 3n + 2
n - 1 + 2n + 2 = 3n + 2
3n + 1 = 3n + 2
0 = 1

no solution.