a) (20y)/(36y^2)-4-(2y-3)/(2-6y)=(5-2y)/(6y+2)

To solve the equation $\frac {20y}{36y^{2}-4}-\frac {2y-3}{2-6y}=\frac {5-2y}{6y+2}$, we need to find a common denominator for the fractions on the left-hand side and then simplify the equation.<br /><br />Step 1: Find a common denominator for the fractions on the left-hand side.<br />The denominators are $36y^2 - 4$ and $2 - 6y$. We can factor out a 4 from the first denominator to get $4(9y^2 - 1)$, and we can factor out a -2 from the second denominator to get $-2(1 - 3y)$. The common denominator is $4(9y^2 - 1)(-2(1 - 3y))$.<br /><br />Step 2: Rewrite the fractions with the common denominator.<br />$\frac {20y}{36y^{2}-4}-\frac {2y-3}{2-6y}=\frac {5-2y}{6y+2}$<br />$\frac {20y(-2(1 - 3y))}{4(9y^2 - 1)(-2(1 - 3y))}-\frac {(2y-3)4(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {(5-2y)4(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br /><br />Step 3: Simplify the numerators.<br />$\frac {-40y(1 - 3y)}{4(9y^2 - 1)(-2(1 - 3y))}-\frac {(2y-3)4(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {(5-2y)4(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br />$\frac {-40y + 120y^2}{4(9y^2 - 1)(-2(1 - 3y))}-\frac {4(2y-3)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br /><br />Step 4: Combine the fractions on the left-hand side.<br />$\frac {-40y + 120y^2 - 4(2y-3)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br /><br />Step 5: Simplify the numerator on the left-hand side.<br />$\frac {-40y + 120y^2 - 4(2y-3)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br />$\frac {-40y + 120y^2 - 4(18y^3 - 12y^2 - 27y + 3)}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br />$\frac {-40y + 120y^2 - 72y^3 + 48y^2 + 108y - 12}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 - 1)}{4(9y^2 - 1)(-2(1 - 3y))}$<br />$\frac {-72y^3 + 168y^2 + 68y - 12}{4(9y^2 - 1)(-2(1 - 3y))}=\frac {4(5-2y)(9y^2 -