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3 answers

7(x-7) = 6(x+3) (multiplied both sides by 42)
7x -49 = 6x + 18 (distributive property)
x = 67 (moving x's to one side, constants to the other)

2007-09-27 15:23:46 · answer #1 · answered by lhvinny 7 · 0 0

(x-7)/6 = (x+3)/7
(1/6)*(x-7) = (1/7)*(x+3)
We want to solve the equation (find the value of x that makes the equation true). To do this, we must isolate the value of x onto one side of the equation and get its multiplier equal to 1. We can add/subtract/multiply/divide both sides of an equation by the same amount, and still keep the equality.
First, lets get rid of the fractions.
6*(1/6)*(x-7) = 6*(1/7)*(x+3)
(x-7) = (6/7)*(x+3)
7*(x-7) = 7*(6/7)*(x+3)
7*(x-7) = 6*(x+3)
now lets multiply this out:
7*x - 49 = 6*x + 18
and lets get the x all on one side:
7*x - 49 + 49 = 6*x + 18 + 49
7*x = 6*x + 67
7*x - 6*x = 6*x + 67 - 6*x
1*x = 67
x = 67

now we should check our work. plug x = 67 back into the original equation and see if it works out.
(x-7)/6 = (x+3)/7
(67-7)/6 ?= (67+3)/7
(60)/6 ?= (70)/7
10 = 10
yea, our answer works! :)

2007-09-27 22:28:25 · answer #2 · answered by mrvadeboncoeur 7 · 0 0

(x - 7) / 6 = (x + 3)/ 7 multiply both side by 6
6(x-7)/6 = ((x + 3)/ 7 ) 6

x - 7 = (6x + 18)/7 then multiply both side by 7
7(x-7) = ((6x + 18)/7) 7

7x - 49 = 6x + 18 subtract both side by 6x
- 6x -6x
1x - 49 = 18 then plus 49 for both side
x = 67

Here is the link for your future use.

http://www.algebra.com/

2007-09-27 22:54:12 · answer #3 · answered by tinhmay 1 · 0 0

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