Since the integers are consecutive, you could call them 'x', x+1,and x+2.
Or you could all them x-1, 'x', and x+1.
Or you could call them x-2, x-1, and 'x'.
It's up to you. Each of those sets describes 3 consecutive numbers.
Most people will call them 'x', x+1, and x+2, so let's do that.
Now just go through the question and make the quantities it talks about:
"twice the first" . . . . 2(x)
"three times the second" . . . . 3(x+1)
"four times the third" . . . . . 4(x+2)
and finally, it says that the sum of those things is 20.
This should not be a major problem.
2x + 3(x+1) + 4(x+2) = 20
You can probably handle it from here, but I'll continue anyway.
The next thing to do is take out all the parentheses:
3(x+1) = 3x + 3
4(x+2) = 4x + 8
Now add up the whole thing again, but without parentheses:
2x + 3x + 3 + 4x + 8 = 20
Gather all the x's together, and all the naked numbers together:
(2x + 3x + 4x) + (3 + 8) = 20
Addum up: 9x + 11 = 20
Subtract 11 from each side: 9x = 9
Divide each side by 9 : x = 1
The first integer = x . . . . . . . 1
The second integer = x+1 . . . 2
The third integer = x+2 . . . . . 3
Check:
"twice the first" . . . . . . . . . . 2(1) = 2
"three times the second". . . 3(2) = 6
"four times the third" . . . . . . 4(3) = 12
Addum up: 2 + 6 + 12 = 20 just like the question said. yay!
