Some Questions on Algebra (Secondary1)?

Question

Q1...........3(4(5a-3a)-5(4a-a))-9a
Q2...........4(2x-y)-5(2x-3(x-2y))
Q3........... 12x
-------------------------
5x+3((4x-8x)+x)

Q4...........(3a)(2de)(def)
Q5...........25c X 25c X 25c X 25c X 25c X 25c + d X d X d X d
Q6...........x(3x+2)-(2x+4)

how clear???????/wat ques?????

srry ....

q1 : 3(4(5a-3a)-5(4a-a))-9a

q2: 4(2x-y)-5(2x-3(x-2y))

Answer 1

1) 3{4(5a-3a)-5(4a-a)}-9a
=3{4*2a-5*3a)-9a
=24a-45a-9a
= -30a
2)4(2x-y)-5{2x-3(x-2y)}
=8x-4y-5(2x-3x+6y)
=8x-4y-5(-x+6y)
=8x-4y+5x-30y
=13x-34y
3) Denominator
=5x+3(4x-8x)+x
=5x+12x-24x+x
= -6x
Therefore Given expression
=12x/-6x
= -2
4)(3a)(2de)(def)
=6ad^2e^2f
5)25c*25c*25c*25c*25c*25c-d*d*d*d*
=(25c)^6-d^4
6) x(3x+2)-(2x+4)
=3x^2+2x-2x-4
=3x^2-4

Answer 2

Not really sure of your notation on the first 2 quesitons, so here are the answers to the last 4

q3.
expand the bottom line first
5x+3(-3x)
=5x-9x
=-4x

so you have 12x/-4x = -3


Question 4.

Do the first two brackets first
6ade (def)
then do the last
6ad^2e^2f

Question 5
I would write it as
25^6c^6 + d^4

Question 6
3x^2+2x-2x-4
=3x^2-4

Answer 3

please write your questions clearly and correctly.