9/5(3-x)=3/4(x-3)? How do you solve this? is the L.C.M. 20?
回答 (9)
2009-05-25 3:53 pm
✔ 最佳答案
9/5(3-x)=3/4(x-3) lcm:2036(3-x)=15(x-3)
108-36x=15x-45
-36x-15x=-45-108
-51x=-153
x=3 answer//
2009-05-28 7:36 pm
Incorrectly presented.
Should be clearly shown as :-
(9/5) ( 3 - x ) = (3/4) ( x - 3 )
36 ( 3 - x ) = 15 ( x - 3 )
108 - 36 x = 15 x - 45
153 = 51 x
x = 3
Should be clearly shown as :-
(9/5) ( 3 - x ) = (3/4) ( x - 3 )
36 ( 3 - x ) = 15 ( x - 3 )
108 - 36 x = 15 x - 45
153 = 51 x
x = 3
2009-05-25 4:20 pm
5 & 4 have not common factor, so LCM = 5 x 4 = 20
9/5(3-x)=3/4(x-3)
factor out -1 from (x -3)
9/5(3-x)=3/- 4(3- x)
9/5(3-x) + 3/4(3-x) = 0
9/5(3-x)=3/4(x-3)
factor out -1 from (x -3)
9/5(3-x)=3/- 4(3- x)
9/5(3-x) + 3/4(3-x) = 0
2009-05-25 4:14 pm
9/5(3 - x) = 3/4(x - 3)
27/5 - 9/5x = 3/4x - 9/4
- 9/5x - 3/4x = - 27/5 - 9/4
- 36/20x - 15/20x = - 108/20 - 45/20
- 51/20x = - 153/20
(20/51)-51/20x = (20/51)-153/20
- x = - 3
or
x = 3.
27/5 - 9/5x = 3/4x - 9/4
- 9/5x - 3/4x = - 27/5 - 9/4
- 36/20x - 15/20x = - 108/20 - 45/20
- 51/20x = - 153/20
(20/51)-51/20x = (20/51)-153/20
- x = - 3
or
x = 3.
2009-05-25 4:10 pm
9/5(3 - x) = 3/4(x - 3)
[9(3 - x)]/5 = [3(x - 3)]/4
(27 - 9x)/5 = (3x - 9)/4
4(27 - 9x) = 5(3x - 9)
108 - 36x = 15x - 45
-15x - 36x = -108 - 45
-51x = -153
x = -153/-51
x = 3
[9(3 - x)]/5 = [3(x - 3)]/4
(27 - 9x)/5 = (3x - 9)/4
4(27 - 9x) = 5(3x - 9)
108 - 36x = 15x - 45
-15x - 36x = -108 - 45
-51x = -153
x = -153/-51
x = 3
2009-05-25 4:00 pm
x= 3
2009-05-25 3:56 pm
Here's a more creative approach:
Factor a (-1) out of the left parenthesis. You get -9/5 (x-3) = 3/4 (x-3)
Clearly -9/5 never equals 3/4.
Therefore (x-3) must equal zero.
Then we have x=3.
You can check it in the original equation, and you have 0=0, which is true.
Factor a (-1) out of the left parenthesis. You get -9/5 (x-3) = 3/4 (x-3)
Clearly -9/5 never equals 3/4.
Therefore (x-3) must equal zero.
Then we have x=3.
You can check it in the original equation, and you have 0=0, which is true.
2009-05-25 3:50 pm
The LCM of the denominators is 20, since 5 and 4 have no common factors. Let's multiply both sides by 20:
20* 9/5(3-x)= 20* 3/4(x-3)
Using cross canceling on the fractions, we get
36(3-x) = 15(x-3).
Now, use the distributive property on each side:
36 * 3 - 36x = 15x - 15 * 3
108 -36x = 15x -45.
To solve for x, we need to get all the x's on one side of the equation. The most convenient way is to add 36x to both sides:
108 = (15x - 45) + 36x
or
108 = 51 x - 45.
Now, this is easy; add 45 to both sides:
153 = 51 x
and divide both sides by 51:
x = 3.
20* 9/5(3-x)= 20* 3/4(x-3)
Using cross canceling on the fractions, we get
36(3-x) = 15(x-3).
Now, use the distributive property on each side:
36 * 3 - 36x = 15x - 15 * 3
108 -36x = 15x -45.
To solve for x, we need to get all the x's on one side of the equation. The most convenient way is to add 36x to both sides:
108 = (15x - 45) + 36x
or
108 = 51 x - 45.
Now, this is easy; add 45 to both sides:
153 = 51 x
and divide both sides by 51:
x = 3.
2009-05-25 3:46 pm
You can use lcm
36/20(3-x) = 15/20(x-3)
multiply bs by 20
36(3-x) = 15(x-3) = -15(3-x) Can you guess what is coming?
(36+15)(3-x) = 0
51(3-x) = 0
so 3-x must equal 0, x=3
36/20(3-x) = 15/20(x-3)
multiply bs by 20
36(3-x) = 15(x-3) = -15(3-x) Can you guess what is coming?
(36+15)(3-x) = 0
51(3-x) = 0
so 3-x must equal 0, x=3
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