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l0rrie
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Thanks.. I also tried plugging in numbers but I was wondering if someone here could explain the algebraic approach to me.. Thanks a lot


5 (x-1)+ 420/x=90
x=12

p.s.-in my 1st post I actually tried to explain. to make it clear I began from the end ;) and then wrote the formula
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I got the numbers right but I think I did something wrong in my calculation.. I can't solve for x I keep getting x is 4,4 when Im trying to solve this equation.. Could you please write it down for me how you solved for x is 12, thanks a bunch!
Here's what I got:

420/x + 5x = 95x
420 = 90x
x=4,4 :S
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l0rrie
Mrs. Anderson divided $420 evenly among her children. Later, when Mrs. Anderson was away, Daniel, the eldest, took $5 from each of his brothers and sisters. Daniel ended up with $90. If Mrs. Anderson has more than 7 children, how many children does Mrs. Anderson have?

12
14
16
18
20


Here's what I did:

420/x + 5(x-1) = 90

However, I can't seem to solve for x.. Can somebody please help me :(

\(\frac{420}{x}+5(x-1)=90\)
\(\frac{420}{x}+5x-5=90\)
\(\frac{420+5x^2-5x}{x}=90\)
\(420+5x^2-5x=90x\)
\(5x^2-95x+420=0\)
Divide both sides by 5;
\(x^2-19x+84=0\)
\(x^2-12x-7x+84=0\)
\(x(x-12)-7(x-12)=0\)
\((x-7)(x-12)=0\)

x=7 OR x=12;
We know Mrs. Anderson has more than 7 children;
Thus, x=12.

Ans: "A"
*****************************************

PIN also works good for this.
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Thanks so much fluke, that was exactly what I was looking for.. Why was my approach wrong.. I basically did:

420/x + 5x -5 = 90
420/x + 5x = 95 (multiplied both sides with x)
420 + 5x = 95x
420 = 90x

If you multiply by x that doesn't mean you have to do it for everything on the left side right, so in this case also the 5x? I always thought that you only need to multiply one 'one' part of either side..
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Thanks so much fluke, that was exactly what I was looking for.. Why was my approach wrong.. I basically did:

420/x + 5x -5 = 90
420/x + 5x = 95 (multiplied both sides with x)
420 + 5x = 95x
420 = 90x

If you multiply by x that doesn't mean you have to do it for everything on the left side right, so in this case also the 5x? I always thought that you only need to multiply one 'one' part of either side..

Any operation you do for LHS, you have to do it for RHS.
Also, the operation is not selective for multiplication/division. You have perform the operation for all terms for multiplication/division.

420/x + 5x = 95
Multiply both sides by x;

(420/x)*x+5x*x=95*x
420+5x^2=95x
Subtract 95x from both sides;
420+5x^2-95x=95x-95x
420+5x^2-95x=0
Rearrange:
5x^2-95x+420=0
Divide both sides by 5;
5x^2/5 - 95x/5 + 420/5 = 0/5
x^2-19x+84=0
*******************************
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Again thanks, very clear now.. The rule you just described, is that only for multiplying/dividing?
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Again thanks, very clear now.. The rule you just described, is that only for multiplying/dividing?

Yes, that's only for "multiplication/division" thanks



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