y÷(2y-3) + 4÷(y+1) ...help?

Question

I got this question earlier this week in my Maths GCSE paper. I had no idea how to solve it but I think the solution is 1 so I wrote that, hopefully I'll get some marks. Anyway I was curious if I was right, and how you're actually supposed to solve it.

:: Edit ::

My bad, it is an equation the full question was...

y÷(2y-3) + 4÷(y+1) = 1

It was worth 4 or 5 marks.

Answer 1

y/(2y-3) + 4/(y+1) = 1

y(y+1)/(2y-3)(y+1) + 4(2y-3)/(2y-3)(y+1) = 1

Now multiply through to cancel denominators.

y(y+1) + 4(2y-3) = (2y-3)(y+1)

= y^2 + y + 8y - 12 = 2y^2 + 2y -3y - 3

= -y^2 + 10y - 9

Set to zero, and change signs.

0 = y^2 - 10 y + 9

Factor.

0 = (x-9)(x-1)

Therefore, since this is a quadratic equation, there are two answers (x=9, or x=1)

Hope that helps. Good luck.

Answer 2

y/(2y-3) + 4/(y+1) = 1

y(y+1) + 4(2y-3) = 1(y+1)(2y-3)

y² + y + 8y -12 = 2y² -y -3

y² + 9y -12 = 2y² -y -3

um...

0= y² - 10y +9

factor it out

(y-9)(y-1) = y² - 10y +9

change the signs around to get the correct answer:

y= 9
and/or
y=1

so, you were half right. you were just missing the 9

Answer 3

You know, I can't solve for y unless you give me the sum. If it were y/2y+-3 = 4/y+1, that would be a different story.

Answer 4

y / (2y - 3) + 4 / (y + 1) = 1
[ y.(y + 1) + 4.(2y - 3) ] / (2y - 3).(y + 1) = 1
[ y² + 9y - 12 ] / (2y - 3).(y + 1) = 1
y² + 9y - 12 = 2y² - y - 3
y² - 10y + 9 = 0
(y - 9).(y - 1) = 0
y = 9, y = 1

Answer 5

IN THAT CASE, I believe that you are correct!
I didn't actually figure it out step by step, but I checked your answer in the equation and it worked out.