Maths help needed please. 10 points to first correct anwser?

Multiply both sides by (y+1)(2y-3)
so you get,
2(2y-3)+3(y+1) = (y+1)(2y-3)
7y -3 = 2y^2-y-3,
or
2y^2 - 8y=0. so y^2-4y=0
y(y-4) =0 so y = 0 or 4.
U may cross check by substituting.

2/(y + 1) + 3/(2y - 3) = 1
(2[2y - 3] + 3[y + 1])/([y + 1][2y - 3]) = 1
4y - 6 + 3y + 3 = 2y² - 3y + 2y - 3
2y² - 8y = 0
2y(y - 4) = 0

Answer: y = 0, 4; factors: 2y(y - 4)

Proof (y = 0):
2/(0 + 1) + 3/(2[0] - 3) = 1
2/1 + 3/-3 = 1
2 - 1 = 1
1 = 1

Proof (y = 4):
2/(4 + 1) + 3/(2[4] - 3) = 1
2/5 + 3/(8 - 3) = 1
2/5 + 3/5 = 1
5/5 = 1
1 = 1

2/(y + 1) + 3/(2y - 3) = 1
(y + 1)(2y - 3)[2/(y + 1) + 3/(2y - 3)] = (y + 1)(2y - 3)(1)
2(2y - 3) + 3(y + 1) = (y)(2y) + (y)(-3) + (1)(2y) + (1)(-3)
2(2y) - 2(3) + 3(y) + 3(1) = 2y^2 + (-3y) + 2y + (-3)
4y - 6 + 3y + 3 = 2y^2 - 3y + 2y - 3
2y^2 - y - 3 - 4y + 6 - 3y - 3 = 0
2y^2 - y - 4y - 3y - 3 + 6 - 3 = 0
2y^2 - 8y = 0
y^2 - 4y = 0
y(y - 4) = 0

y = 0

y - 4 = 0
y = 4

∴ y = 0, 4

2/(y+1) + 3(2y-3) = 1
Combining two terms, we have
[2(2y-3) + 3(y+1)]/[(y+1)(2y-3)] = 1
(4y-6+3y+3)/(2y^2 - y -3) = 1
4y-6+3y+3 = 2y^2 - y -3
7y-3 = 2y^2 -y-3
2y^2 -8y = 0
2y(y-4) = 0
y=0 ; y=4
y {0,4}

solve this equation 2(2y-3)+3(y+1)=(y+1)(2y-3)