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The problem says:
"The sum of a number and seven is divided by four.The result is 19 more then the number. Find the number."

How would I set that up? I am not asking what the answer is, I just want to know how I would set that up on my paper.

2006-09-24 06:54:35 · 10 answers · asked by whatisn'twouldn'tbe™ 4 in Education & Reference Homework Help

Thank you all so much!
I have been working on the problem for a while now. It makes more sense now! THANKS!!! :]

2006-09-24 07:09:51 · update #1

10 answers

(x+7)/4=x+19
x+7=4x+76
3x=-69
x=-23

2006-09-24 06:57:10 · answer #1 · answered by raj 7 · 0 0

You can usually write your equation by following the sentences you're given. "The sum of a number and seven" gets put in parentheses because the whole thing is divided by 4.

(x+7)/4 = 19 + x
Multiply both sides by 4 to get rid of the fraction.
x + 7 = 76 + 4x

I'm sure you can take it from there.

2006-09-24 14:01:33 · answer #2 · answered by PatsyBee 4 · 0 0

(x=7)/4 = x+19

2006-09-24 13:58:40 · answer #3 · answered by smitty♥ 2 · 0 0

(x+7)/4 =x+19

2006-09-24 13:57:51 · answer #4 · answered by Yup! I'm a girl! 2 · 0 0

let the no be X
X + 7 /4 = X +19

X+7 = 4(x +19).
x+7 = 4x +76
4x -x = -76 +7
3x = -69
x= -69/3
x= -23

2006-09-24 13:59:21 · answer #5 · answered by macline k 2 · 0 0

(x+7)/4=x+19

2006-09-24 13:56:41 · answer #6 · answered by Anonymous · 1 0

(x + 7)/4 = 19+ x that should be the equation

2006-09-24 13:58:54 · answer #7 · answered by phil b 1 · 0 0

(x+7)/4 = 19 multiply both sides by 4
(x+7) = 76 subtract 7 from both sides
x = 69

2006-09-24 13:59:54 · answer #8 · answered by Grev 4 · 0 0

x+7/4 = x+19
just write it the way it is written in words. say your written words as you read what i wrote.
once you do many of these it will come easier.
math mostly just memorize how to formulas not as hard as it seems..

2006-09-24 14:00:26 · answer #9 · answered by macdoodle 5 · 0 0

Let the number be x
Given (x + 7)/4=19 + x
Hence x = -23 (Verify please)

2006-09-24 13:58:00 · answer #10 · answered by venkatesh p 1 · 0 0

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