1/6(3x+5y)+2/3(3/5x-6y)?

Question

"i was wondering if u could help me with this problem im confused and dont know where to begin

Answer 1

multiply everything out

3x/6 + 5y/6 + 6x/15 - 4y

get common denominators for all (which would be 30)

15x/30 + 25y/30 + 12x/30 - 120y/30

32x -95y
_______
30

that would be the simplified answer

Answer 2

you commence with 4(5y-3) utilising the distributive resources. (4*5y)-(4*3) this provides 20y-12 Now we address (6y+3). The "-" in the front of this component can be appeared at as "-a million," so that you're utilising the distributive resources back (-a million*6y)+(-a million*3) = -6y-3 once you tack it back onto something else, it turns into: 20y-12-6y-3 The associative resources of addition states a+b=b+a, so that you'll rearrange those values (contain the +/- indications to all the values). combine like values: 20y-6y-12-3 20y-6y=14y and -12-3=-15 Your answer is 14y-15 EDIT: i'm not affirming you probably did this, yet one mistakes in searching at this expression is viewing the "-(6y+3)" section as -6y+3. The parentheses teach 6y+3 as one section so it isn't 4(5y-3) - 6y+3, yet 4(5y-3) subtracted through the entire expression 6y+3 ( 4[5y-3] subtract 6y and subtract +3). -(6y+3) would seem as if this: -6y - +3 (or purely -3 to make: -6y-3)

Answer 3

Begin by multiplying everything out. So for example the first half (before the plus sign) is (3/6)x + (5/6)y. Then group the x terms together and y terms together and simplify.

Answer 4

first you multiply whats outside the bracket by whats inside so
1/6=0.166666666*3x=0.5x 0.166666666*5y=0.833333333y
and
2/3=0.666666666*3/5=0.4x 0.666666666*6=4y


so 0.5x+0.833333333y+0.4x-4y

then group the same letters together
0.5x+0.4x=0.9x 0.833333333y-4y= -3.166666667y

0.9x+- 3.166666667y is your answer. there is no total solution to this question!!!!

if you cant be bothered working that out, the calculator will do it for you!!!!!

Answer 5

It has to equal something, or there's nothing you can do with it.

Answer 6

is "x" or "y" given