Question

2√y-1-1√3y-1=0

Answer 1

think there might be a typo here

Answer 2

2√y-1- √3y-1=0

Answer 3

ok then
\[2\sqrt{y-1}-\sqrt{3y-1}=0\] right?

Answer 4

yeah that is right sorry this math has got me crazy

Answer 5

\[2\sqrt{y-1}=\sqrt{3y-1}\]square both sides (carefully) solve for y

Answer 6

\[4(y-1)=3y-1\]
notice you have to square the 2 as well. good from here?

Answer 7

???? these problems for some reason I am lost.

Answer 8

The point is to solve for y. So your goal is to isolate y on one side

Answer 9

Did you get the initial steps satellite did for you?

Answer 10

can u help I dont know how to do that

Answer 11

Okay well to begin with you have:\[2\sqrt{y-1} - \sqrt{3y-1}=0\]

Answer 12

You have to solve for y but in this case y is in the form of a radical. So you have to use the inverse operation which is the exponent to get rid of the radical. To make it easier we will move -sqrt(3y-1) to the right side of the equation and that makes it positive

Answer 13

\[2\sqrt{y-1}=\sqrt{3y-1}\]

Answer 14

Do you get it so far. The number one rule for solving equations is that whatever you apply on one side of the equal side must also be applied on the other side of the equal sign that is why we moved the other radical

Answer 15

To get rid of the square roots we square both sides of the equation. Because there is a 2 on the left side that also becomes 4.\[4(y-1)=3y-1\]

Answer 16

Now we will remove the parentheses by multiplying the 4 by the y and -1 to get\[4y-4=3y-1\]

Answer 17

Now there are two ways of going about this. Conventionally we solve for the variable on the left so what we can do first is subtract 3y from both sides of the equation. And then we will also add 4 to both sides of the equation\[4y - 3y -4 + 4 = 3y - 3y -1 + 4\]