Algebra Help Involving Consecutive integers.?

Question

How do i put this into an equation? Find three consecutive integers such that the sum of the first and twice the second is 17 more than twice the third.

Answer 1

Well, here is what you already know

Call the number "X".

x + (x+1) + (x+2) is how you write the three consecutive integers.
That is to say, a number, and a number plus 1, and a number plus 2.

So lets call these as follows

x = x
x+1 = y
x+2=z

so you can now write the whole equation as

(x + 2y) = 2z + 17
This equation is spoken as

The sum of the first (x) and 2 times the second (2y) is equal to 2 times the third plus 17 (2x + 17)

And since we know how each number compares to the other, we can re-write this equation and substitute the values. So it becomes

x = x
x+1 = y
x+2=z

(x + 2y) = 2z + 17
re-write and substitute
x + 2(x+1) = 2(x+2) + 17
Solve
x + 2x+2 = 2x+4 + 17
3x + 2 = 2x + 21

x = 19
That means that y = x+1 = 20
and z = x+2 = 21

Prove
(x + 2y) = 2z + 17
19 + 2(20) = 2(21) + 17
19 + 40 = 42 + 17
59 = 59 proved

Answer 2

Just to write an equation you could use:

let x, y, and z be three consecutive integers

x+2y-17=2z
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or, if you want to find what the integers are then...

let x, x+1, and x+2 be three consecutive integers

x + 2(x+1)-17 = 2(x+2)

x + 2x + 2 - 17 = 2x + 4
3x -15 = 2x + 4
x = 19

so, the integers are 19, 20, and 21

Answer 3

Let x, x+1, and x+2 represent the consecutive integers.

x + 2(x+1) = 2(x+2) +17

x + 2x +2 = 2x + 4 +17
3x + 2 = 2x + 21
x = 19

Sub back in to find the other two: 19,20, 21

Answer 4

x + 2(x+1) = 2(x+2) + 17

x + 2x + 2 = 2x + 4 + 17

3x +2 = 2x + 21 ; subtract 2x and 2 from both sides.

x = 19.

check real quick

19 + 40 = 42 + 17
59 = 59

Answer 5

x+2(x+1)=2(x+2)+17

x+2x+2=2x+4+17

3x+2=2x+21
-2 -2
3x=2x+19
-2x-2x
1x=19
or
x=19

then you isnert for x and get 59=59!