Can I get some assistance with clearing out fractions in a linear equation, or tell me if solved it correctly?

Question

My problem is 7/3(2d-1)-2/5(4-3d)=1/5d

I know my LCD is going to be 15, so, would it be set up like this...

15*7/3(15*2d-15*1)-15*2/5(15*4-15*3d)=15*1/5d

5*7(30d-15)-6(60-45d)=3d

here is where I get stuck...I forgot if I distribute the 7 first then distribute the 5

thanks...

oh, I forgot to take out the "tell me if I solved it correctly" because I know my answer didn't make sense, therefore, I don't have an answer listed.

Thanks so far to those that have answered, now I understand where I was going wrong.

Answer 1

7/3(2d-1)-2/5(4-3d)=1/5d

14/3d - 7/3 -8/5 +6/5d =1/5d

14/3d +6/5d -1/5d = 7/3 +8/5

[70d+ 18d -3d] /15 = [35 +24] /15

70d+ 18d -3d = 35+24
85d = 59
d=59/85

Answer 2

Brackets are required to avoid confusion:-
(7/3).(2d - 1) - (2/5).(4 - 3d) = (1/5).d
Multiply by 15:-
35.(2d - 1) - 6(4 - 3d) = 3d
70d - 35 - 24 + 18d = 3d
85d = 59
d = 59/85

Answer 3

7/3(2d -1) - 2/5(4-3d) = 1/5d

Okay, let's multiply both sides by 15

15( 7/3 (2d-1) - 2/5(4-3d) = 1/5d)

15*7/3(2d-1) - 15*2/5(4-3d) = 15*1/5d
105/3(2d-1) - 30/5(4-3d) = 15/5d
35(2d-1) - 6(4-3d) = 3d
70d - 35 - 24 + 18d = 3d
88d -59 = 3d
88d - 3d = 59
85d = 59
d = 59/85

Answer 4

(7/3)*(2d-1)-(2/5)*(4-3d)=1/(5d)
Let a = 2d-1, b=4-3d,
15*(7/3)*a - 15*(2/5)*b = 1*15/5d
5*7a-3*2b = 3/d
35a-6b = 3/d

multiply both sides by d

35d*a-6d*b = 3
substitute for a and b

35d*(2d-1)-6d*(4-3d)=3

Expanding this gives

70d^2-35d-24d+18d^2=3

88d^2-59d=3

Using the quadratic equation for solving for d you can find the values for d

Answer 5

Okeedokee :) What you prefer to do in this situation is make it so as that y = stuff, yet y is the only element on the left area, so... y/6 + 4 = 7 y/6 = 3 Subtract 4 from the two facets y = 18 Multiply the two facets via 6 So your answer is y = 18