How do you solve this (3y+1)^1/2=y-1

y=-3

Step-by-step explanation:

(number)^1 is just (number)

so (3y+1)^1/2=y-1 is (3y+1)/2=y-1

multiply 2 on both side so get rid of fraction

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

3y+1=2y-2

y=-3

it's equivalent, so we can say that its equal.

let's call the denominator x and the numerador is 18 less then denominator, x - 18.

multiply

2.x = 5.(x - 18)

2x = 5x - 90

90 = 5x - 2x

90 = 3x

x = 90/3

x = 30

so: