Question

Solve

Answer 1

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

Answer 2

@cshalvey

Answer 3

@.UserNotFound. @jazy @cshalvey

Answer 4

@KingGeorge

Answer 5

In the future, please refrain from tagging a bunch of users.

Anyways, first move the 3 over.\[\sqrt{2y-3}+3=y\]\[\sqrt{2y-3}=y-3\]Now square both sides. Can you tell me what youget?

Answer 6

sorry, they're all the people that i have fanned for helping me.

Answer 7

No worries, just something to remember for the future. Anyway, can you tell me what you get after squaring both sides?

Answer 8

\[\left( \sqrt{2y-3} \right)^{2}=\left( y+3 \right)^{2}\]
\[\left( 2y-3 \right)=\left(y ^{2}+9 \right)\]

Answer 9

The left side looks good. The right is not correct. Remember that you need to FOIL. Can you try again?

Answer 10

whats a "FOIL"?

Answer 11

It's an acronym for

First
Outer
Inner
Last

It means you first multiply the first two numbers together, then the outer numbers together, then the inner, and then the last numbers. Once you've done that, you add them all. For example:

\[(y+3)(y+2)\]
First: \((\color{red}y+3)(\color{red}y+2)\implies y\cdot y=y^2\)
Outer: \((\color{red}y+3)(y+\color{red}2)\implies y\cdot 2=2y\)
Inner: \((y+\color{red}3)(\color{red}y+2)\implies 3\cdot y=3y\)
Last: \((y+\color{red}3)(y+\color{red}2)\implies 3\cdot2=6\)

So \((y+3)(y+2)=y^2+2y+3y+6=y^2+5y+6\)

Answer 12

Now can you try the same thing with \((y+3)^2=(y+3)(y+3)\)?

Answer 13

Oh, and it should be \((y-3)^2=(y-3)(y-3)\). Oops :(

Answer 14

\[\left( y+3 \right)\left( y+3 \right)\]
\[y^{2}+3y+3y+9?\]

Answer 15

Right, and if you had \((y-3)(y-3)\), it would become something similar. You get \[y^2-3y-3y+9=y^2-6y+9\]
This means you have \[2y-3=y^2-6y+9\]Get 0 on one side, and you get \[y^2-8y+12=0\]Now you just need to factor this. Can you manage that yourself?

Answer 16

yes i can. Thank you soo much king. for every thing. ima fan you

Answer 17

You're welcome. Feel free to post what you get as solutions and I can check them if you want.

Answer 18

kk will do. Thanks