Maths Equation

7[2(y-3)+y]=4[1-3(y+2)]
7(2y-6+y) = 4(1-3y-6)
7(3y-6) = 4(-3y-5)
21y - 42 = -12y - 20
33y = 22
y = 2/3

To solve this kind of equation, you must first expand all the terms, that is, remove the brackets. It is generally easier to consider the small brackets first.
7[2(y-3) + y] = 4[1 - 3(y+2)]
7[2y - 6 + y] = 4[1 - 3y - 6]
14y - 42 + 7y = 4 - 12y - 24

After this we should add the similar terms together. It becomes ...
21y - 42 = -20 - 12y

Then we put the unknown terms (terms with y) in one side and constant terms on the other side. It becomes ...
21y + 12y = -20 + 42
33y = 22

Finally, we divide the constant with the coefficient of the unknown term.
33y = 22
33y/33 = 22/33
y = 22/33 = 2/3

7[2(y-3)+y]=4[1-3(y+2)]
7(2y-6+y) = 4(1-3y-6)
7(3y-6)=4(-5-3y)
21y-42= -20-12y
33y=22
y=22/33
y=2/3

7[2(y-3)+y]=4[1-3(y+2)]
14(y-3)+7y=4-12(y+2)
14y-42+7y=4-12y-24
21y-42=-20-12y
21y+12y=-20+42
33y=22
y=2/3

7[2(y-3)+y]=4[1-3(y+2)]
7(2y-6+y)=4(1-3y-6)
7(3y-6)=4(-5-3y)
21y-42=-20-12y
33y=22
y=2/3

7(2y-6+y)=4(1-3y-6)
21y-42=-20-12y
33y=22
y=2/3