SOLUTION: I need help with this question, cant seem to get the right answer. A rectangle with an area of 80cm2 has a width that is three more than one half its length. What is the width o

Question 299944: I need help with this question, cant seem to get the right answer.
A rectangle with an area of 80cm2 has a width that is three more than one half its length. What is the width of the rectangle
Found 2 solutions by JBarnum, wise1:
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
{{[A=80}}}

2 numbers multiply to get -160 add to get +6: -10 and 16

, cant be negative

Answer by wise1(2) (Show Source): You can put this solution on YOUR website!
before u go through all this, pls read the last 4 lines
area=80cm2,
to make things easier, lets call length 'b' and width 'w'

area=b*w=80
substituting 'w' with

multiply through by 2 to get rid of the fraction

now i will equate everything to zero
......(i)
multiply the first figure(the one with the square to the number without any letter(dont forget the signs)in this case....

now this is the tricky part. from (i) u need to find 2 numbers whose multiple gives you and whose sum gives you 6b.
in this case, its 16b and -10b
now all we have to do is replace 6b with this sum, after all, its all the same
****dont give up, this only looks bulky cos im tryin to explain clearly***

watch closely. this is how to contract. it doesn't matter how its arranged, just take the first 2 no.s and use what they have in common and do the next 2...

b(b+16)-10(b+16)=0
(b+16)(b-10)=0
b+16=0 or b-10=0
b=-16 (not possible) b=10
therefore, b=10
....this might be your problem....were looking for w, not b
w=(b/2)+3
w=(10/2)+3
w=5+3 =8
voila!
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