Math problem...need help!?

Question

Find three consecutive integers such that 6 times the second decreased by twice the first is the same as 14 more than three times the third. I need help solving this, I am confused...I also need to do the check.

Answer 1

x, x+2 and x+4

6(x+2) - 2x = 3(x+4) + 14

6x + 12 - 2x = 3x + 12 + 14

4x + 12 = 3x + 26

x = 14

14,16 and 18

Answer 2

To start of your three consecutive integers are going to be:
n , n+1, and n+2

6 times the second decreased by twice the first is:

6(n+1) - 2n

14 more than 3 times the third is:

14+3(n+2)

and the equal each other so your equation is:

6(n+1)-2n = 14+3(n+2)

6n+6-2n = 14+3n+6

4n+6 = 3n+20

n=14

So then: n=14, n+1=15, and n+2= 16

To check just plug the numbers into the equation:

6(n+1)-2n=14+3(n+2)

6(15)-2(14)=14+3(16)
62=62

***I was looking at the first answer to your question and the guy didn't do three consecutive integers. He set you up to do 3 odd or even integers***

Answer 3

Let the lowest integer be n. (1)
Then the next two are n+1 and n+2. (2)
6 times the second is 6(n+1).
Twice the first is 2n
14 more than 3 times the third is 3(n+2) + 14.
The whole equation therefore is:
6(n+1) - 2n = 3(n+2) + 14 (3)
Multiply out the brackets:
6n + 6 - 2n = 3n + 6 + 14
You now need to get all the n terms on the left and all the numbers on the right.
Take 6 off each side.
6n - 2n = 3n + 14
Take 3n off each side:
6n - 2n - 3n = 14
Combine the terms:
(6 - 2 - 3)n = 14
n = 14.
From (1) and (2) the integers are therefore 14, 15, 16.
To check, put these numbers into each side of (3) in turn:
LHS = 6*15 - 2*14 = 62
RHS = 3*16 + 14 = 62
It checks out.
I'd say Somebody has made a mistake or two.

Answer 4

Since we need three consecutive integers, let them be (x-1), x, and (x+1).

Now, 6 times the second decreased by twice the first is the same as 14 more than three times the third.

So, 6 times (x) - twice (x-1) = 14 + three times (x+1)
6x-2(x-1)=14+3(x+1)
6x-2x+2=14+3x+3
x=15

So, the numbers are 14, 15, and 16.

You can do the check by plugging the numbers back into the initial condition.

6X15-2X14=14+3X16

Answer 5

Three consecutive integers are X, X+1 and X+2

The equation would be 6(X+1)-2X = 3(X+2) + 14
4x+6 = 3X + 20
X = 14

The numbers would be 14, 15, 16

Check 6*15 - 2*14 = 3*16 + 14
90 - 28 = 48 +14
62 = 62

Answer 6

Somebody is wrong, that's not consecutive.

your three numbers are x, x+1 and x+2

6(x+1) - 2x = 3(x+2) +14

6x + 6 - 2x = 3x + 6 + 14

4x + 6 = 3x + 20

x = 14

Thus, your numbers are 14, 15, 16.

Answer 7

first is x
second is y
third is z

x + 1 = y
x + 2 = z

6y-2x = 14 + 3z

is a good start. You can solve the equations using substitution.