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\[\sqrt{2y-3}+3=y\]
@.UserNotFound. @jazy @cshalvey

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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?
sorry, they're all the people that i have fanned for helping me.
No worries, just something to remember for the future. Anyway, can you tell me what you get after squaring both sides?
\[\left( \sqrt{2y-3} \right)^{2}=\left( y+3 \right)^{2}\] \[\left( 2y-3 \right)=\left(y ^{2}+9 \right)\]
The left side looks good. The right is not correct. Remember that you need to FOIL. Can you try again?
whats a "FOIL"?
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 \((y+3)(y+2)=y^2+2y+3y+6=y^2+5y+6\)
Now can you try the same thing with \((y+3)^2=(y+3)(y+3)\)?
Oh, and it should be \((y-3)^2=(y-3)(y-3)\). Oops :(
\[\left( y+3 \right)\left( y+3 \right)\] \[y^{2}+3y+3y+9?\]
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?
yes i can. Thank you soo much king. for every thing. ima fan you
You're welcome. Feel free to post what you get as solutions and I can check them if you want.
kk will do. Thanks

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