English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

what's the equation for this word problem?

The product of twice a number and three is the same as 6 times the number and 3/4. find the number

I thought the equation was 2n(3)=6n-3/4 am I right?


and for the second equation:

If the difference of a number and four is doubled, the result is 1/4 less than the number. Find the number

my equation was 2(n-4)=n-1/4 am I right?

2007-09-07 19:16:21 · 4 answers · asked by sweetbabygirlof1983 2 in Science & Mathematics Mathematics

4 answers

The first one is kinda ambiguous. What you have may well be correct, or it could also be
2(n+3) = 6(n + 3/4)

But the 2'nd you nailed. Good job ☺

Doug

2007-09-07 19:32:38 · answer #1 · answered by doug_donaghue 7 · 0 0

Let n = the number.

I'd rather simplify it the way you presented the problem:
Twice n and 3 = 2n + 3; 6 times n and 3/4 = 6n + 3/4 being the two presentation are equal so we can arrive at the following equation.

2n + 3 = 6n + 3/4
4n = 2 1/4
n = (2 1/4) / 4
n = 9/16

Answer: n = 9/16

Proof (plug 9/16 in lieu of n in the above original equation):
2(9/16) + 3 = 6(9/16) + 3/4
9/8 + 3 = 27/8 + 3/4
9/8 + 24/8 = 27/8 + 6/8
33/8 = 33/8
4 1/8 = 4 1/8

2007-09-08 02:44:51 · answer #2 · answered by Jun Agruda 7 · 3 0

According to the wording you will get

6n = 6n +3/4
which gives
0 = 3/4
that doesn't work
another interpretation is
6n = 6n(3/4)
24n = 18n
which is also impossible!
One last possibility is this badly worded problem ment:
2n+3 = 6n + 3/4
8n + 12 = 24n + 3
9 = 16n
n = 9/16


On the second one you are right
2(n-4)=n-1/4

2n-8 = n -1/4
8n - 32 = 4n -1
4n = 31
n = 31/4

2007-09-08 02:45:45 · answer #3 · answered by 037 G 6 · 0 0

2(x -- 4) = x -- 1/4
=> x = 8 -- 1/4 = 31/4 answer Q(2)
confirmation 2(31/4 -- 4) = 15/2 = 31/4 -- 1/4
2x + 3 = 6x + 3/4
=> 4x = 3 -- 3/4 = 9/4
=> x = 9/16 answer Q (1)
confirmation
2(9/16) + 3 = 33/8 = 6(9/16) + 3/4

2007-09-08 02:33:32 · answer #4 · answered by sv 7 · 0 0

fedest.com, questions and answers