Question

Find 3 consecutive integers such that twice the first, plus three times the second, plus four times the third is 20. Please show work, explain throughly so I can understand what to do for next time please! Thank you sooo much!

Answer 1

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!