Im Stuck

tdrake

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A number X is five more than two times X minus three, whats the value of X and how do I show the work?

I think it would look like this X=(2X-3)+5
 
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i would do it as (please correct me if im wrong) 2x+3 =x-5 therefore you would subtract x from each side leaving you with x+3= -5 then subtract 3 from both sides giving you the answer of x= -8
 
a number x is five more than two times x minus three, whats the value of x and how do i show the work?

I think it would look like this x=(2x-3)+5

x = (2x - 3) + 5

x = 2x - 3 + 5

x = 2x + 2

x = -2

Check: put x = -2 into the right hand side of the given equation

(2x - 3) + 5 = [2*(-2)-3] + 5 = [-4 -3] + 5 = -7 + 5 = -2 = x ....... given equation hence checks.
 
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A number X is five more than two times X minus three, whats the value of X and how do I show the work?

I think it would look like this X=(2X-3)+5

Looks good to me but, if that is the way the problem is stated and the answer given by Subhotosh Khan is incorrect, then you might try
x = 2 (x-3) + 5
It depends on just how you interpret that 'two times X minus three'. Is it (two times x) minus (three) as you wrote or is it (two) times (x minus three) as written above. I would be inclined to the (two times x) minus (three).
 
This would not be a typical algebra problem: "X = (2X - 3) + 5"

But this would be: "X = 2(X - 3) + 5"


If it is the latter case, then it can be unambiguously phrased as:

"A number X is five more than two times the difference X minus three."

or


"A number X is five more than two times the quantity X minus three."
 
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Is there any particular reason we're resurrecting a thread from 2011? :shock:
 
Is there any particular reason we're resurrecting a thread from 2011? :shock:

Because the sum of the digits of 2011 is divisible by the first number in the number AND the first number in the number 2011 is congruent to itself mod the square of the number of the answer of the problem. You (should) have to admit that makes it a pretty special post.
 
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