Maths

1.Make y the subject of the formula x=a^2+1/3y

x=a^2+1/3y
x-a^2=1/3y
y= (x-a^2) / (1/3)
y= 3(x-a^2)

2.Expand (x+2y)^2

(x+2y)(x+2y)
= x^2+2xy+2xy+4y^2
= x^2+4xy+4y^2

or
by identity we know (a+b)^2 = a^2+2ab+b^2 so that

(x+2y)^2 = x^2+(x)(2y)+(2y)^2 = x^2+4xy+4y^2

3.Expand(x+2y)(x-2y)

by identity we know than a^2 - b^2 = (a+b)(a-b) so that

(x+2y)(x-2y)
= x^2-(2y)^2
= x^2 - 4y^2

4.Simplify 3n/m^2-mn-6/2m-2n

Is that 3n/m^2 - mn - 6/(2m-2n) or what? Please clarify


5.Find the LCM of m^2-mn and 2m-2n

m^2-mn = m(m-n)
2m-2n = 2(m-n)

L.C.M. of them = 2m(m-n) or 2m^2 - 2n


6.Factorize m^2-mn

m^2-mn = m(m-n)

7.Factorize 2m-2n

2m-2n = 2(m-n)

1.x=a^2+1/3y [ y ]
x-a^2=1/3y
1/x-a^2=3y
y=1/3(x-a^2)
y=1/3x-3a^2

2.(x+2y)^2
=(x+2y)(x+2y)
=x^2+2xy+2xy+4y^2
=x^2+4xy+4y^2

3.(x+2y)(x-2y)
=x^2-2xy+2xy-4y^2
=x^2-4y^2

4.3n/m^2-mn-6/2m-2n
=3n/m^2-6/2m-mn-2n
=(6n-6m)/2m^2-mn-2n
=6n-6m+2m^3n-4m^2n
=6(n-m)+2m^2(m-2)

5.the LCM of m^2-mn and 2m-2n
(m^2-mn)(2m-2n)
=2m^3-2m^3n-2m^3n+2mn^2
=2m^3-4m^3n+2mn^2

6.m^2-mn
=m(m-n)

7.2m-2n
=2(m-n)