Subtract. Write your answer in simplest form. 3/(y-3) - 1/(y+1)?

Answer 1

*Find the greatest common denominator which is, (y-3)(y+1)

First: combine each fraction with the missing denominator.

[3(y+1)/(y-3)(y+1)] - [1(y-3)/(y+1)(y-3)]

[3y+3/(y-3)(y+1)] - [y-3/(y+1)(y-3)]

[3y+3/(y-3)(y+1)] - [y-3/(y+1)(y-3)]

Sec: combine the numerators.

[3y+3 - (y-3)]/(y-3)(y+1)

[3y+3 - y- (-3)]/(y-3)(y+1)

[3y+3 - y + 3]/(y-3)(y+1)

Solution = [2y+6]/(y-3)(y+1)

Answer 2

Multiply the first term by (y+1)/(y+1) and the second term by
(y-3)/(y-3). This will create a lowest common denominator of (y+1)(y-3)

Now, multiplying in the numerator, you get: (3y+3)/(y-3)(y+1)-(y-3)/(y-3)(y+1) or (3y+3-y+3)/(y-3)(y+1) = 2y+6/(y-3)(y+1) or
2(y+3)/(y-3)(y+1)

Answer 3

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

Answer 4

you have to find common denominator which would be (y-3)(y+1);
Multiply the numerator & denominator of first fraction by y+1 & then the numerator & denominator of 2nd fraction by y-3
You end up w/ 3(y+1)-1(y-3)/(y-3)(y+1) which gives you
3y+3-y+3/(y-3)(y+1) combine the like terms of numerator
2y+6/(y-3)(y+1)

Answer 5

= [3.(y +1) - 1 (y - 3 ) ] / [(y - 3).(y + 1) ]
= [ 3y + 3 - y + 3 ] / [ (y - 3).(y + 1)]
= [ 2y + 6 ] / [(y - 3).(y + 1) ]
= 2 (y + 3) / [ (y - 3).(y + 1) ]

Answer 6

LCD(y - 3 and y+1) = (y-3)(y+1)
So,
3/(y-3) - 1/(y+1) =
3(y+1) - 1(y - 3) =
3y + 3 - y +3 =
2y = -6
y = -6 : 2
y = -3
Solution: {y belongs to R| y = -3}
:>:

Answer 7

3/(y-3)-1/(y+1)

lcd is y^2-2y-3

3(y+1)-1(y-3)
3y+3-y+3
2y+6/y^2-2y-3

Answer 8

square of y-4y-9=0

Answer 9

=[3(y+1)-(y-3)]/[(y-3)(y+1)]
=(2y+6)/[(y-3)(y+1)

Answer 10

3/(y-3)-1/(y+1)
3(y+1)-1(y-3)/(y-3)(y+1)
3y+3-y+3/y^2+y-3y-3
2y+6/y^2-2y-3