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Alja [10]
3 years ago
8

Help with this task it's confusing can u explain how you got the answer

Mathematics
1 answer:
Andru [333]3 years ago
7 0
Oh my goodness !  You were doing such an absolutely beautiful job,
as far as you went, but then you ran into some rough road and quit.

You've got the correct expressions for the ages of the three people:

-- Will . . . w
-- Ben . . . w+3
-- Jan . . . 2(w+3)
You slipped up when you expanded Jan's age:  2(w+3) = <u>2w + 6</u> ,
and it was all down hill from there.

Let's do it again, together:

-- Will . . . w
-- Ben . . . w + 3
-- Jan . . . 2w + 6

Total:  (w + w + 2w) + (3 + 6) = 4w + 9

So the equation is:    <em><u>4w + 9 = 41</u></em>
Now you're supposed to solve it.

Subtract 9 from each side:   4w  =  32

Divide each side by 4:        <u>w = 8</u>

-- Will = w . . . . . 8 y.o.
-- Ben = w+3 . . . 11 y.o.
-- Jan = 2(w+3) . . 22 y.o.

When will Jan be twice as old as Will ?
That'll happen in 'x' years.
At that time, Will will be (8+x) and Jan will be (22+x),
and her age will be double Will's age.

22 + x  =  2(8 + x)

22 + x  =  16 + 2x

Subtract 'x' from each side:    22  =  16 + x

Subtract  16  from each side:  <em>  6 = x</em>

<u>Check:</u>

In 6 years, Jan will be (22+6) = 28,
and Will will be (8+6) = 14 .

28 = twice as old as 14.      yay!

Can I make a little suggestion ?
I'm going to make it anyway:

Your problem was neatness.
You were doing great work in that big open space on the sheet, but it
started to get ragged.  When you tried to look back to see if you made
a mistake, you couldn't find it in the mess.
This is not an easy problem, but you definitely know your stuff. 
I think if you keep it a little neater, you're going to sparkle !
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Since the buses travel in opposite directions, the speed at which they distance themselves is the sum of their speeds.
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