Solve linear equations involving fractional constants and muplications?

[匿名]

2008-01-30 2:43 am

Ook im not sure of im doing this question right ..:
1.) 3/4 (2x-1) -1/3 (5-2x)= -55/22

this is wat i started with but not sure

12 (3/4) (2x-1) -12(1/3) (5-2x) =-55/22

KK n another question im not really understanding

2.) 3/4x +4 = 113/24 -2/3 x

Plz help and explain not just answer the q

I would really appreciate this !!!! :):)

回答 (1)

Nowhere I can see

2008-01-30 3:27 am

✔ 最佳答案

1.) When you mulitply by 12, you need to do that for BOTH sides of the equation.
So, 12 * (3/4) (2x-1) -12 * (1/3) (5-2x) = 12 * (-55/22)
Then, 9*(2x-1) - 4*(5-2x) = -30 (Simplify)
Then, 18x - 9 - 20 + 8x = -30 (Simplify)
Then, 26x - 29 = -30 (Simplify)
Then, 26x = -1 (+29 on BOTH sides)
Finally, x = -1/26 (divide 26 on BOTH sides)

2.) 3/4x +4 = 113/24 -2/3 x
3x + 4*(4) = 4*(113/24) - 4*(2/3)x (multiply by 4 on both sides)
3x + 16 = 113/6 - 8/3x (simplify)
3x + 8/3x = 113/6 - 16 (+8/3x on both sides, and then -16 on both sides)
(17/3)x = 17/6 (simplify)
x = (17/6) * (3/17) (multiply by 3/17 on both sides)
x = 1/2

Again, the idea is to isolate x so you will get x = something.
Whenever you try to cancel terms, you need to do it to BOTH sides of the equation, otherwise it would be like putting weigh on only one side of a balanced scale.

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