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Wait why do I subtract 1 from both sides?!

because they want to know what \(\large \boxed{x}-1\) is equivalent to

So we just put a -1 on both sides? That makes no sense...

you start off with \(\large \boxed{x}\) and do that operation to get \(\large \boxed{x}-1\)

But there is no -1 originally. Where did it come from?

I'm using the subtraction property of equality
if a = b, then a-c = b-c

Ohh okay. And why is there a box around the x on the left?

because that's how they set up the notation in the problem

Does it mean anything?

it's just their fancy way of defining a function

they could have easily said "let y = ..." or "let f(x) = ..."

why they chose a box, who knows

Okay. Can you please give me a similar problem to solve?

I need to be able to do these on my own.

ok what is \[\Large \boxed{x} +2x\] equivalent to?
(same function from the previous problem)

I'm not sure :(

replace the box x with what it is equal to
then simplify

4/x-1 ?

box x was equal to (x+3)/(x-1)

so saying \[\Large \boxed{x}+2x\]is the same as saying (x+3)/(x-1) + 2x. Simplify from here

What would the first step be?

you need to get the denominators the same

2x = 2x/1 has a denominator of 1

how can you transform that 1 into (x-1)

So (x+3)/(x-1) + 2x/1 ?

how can you get 2x/1 to have a denominator of x-1

Multiply 1 on both sides?

why not multiply top and bottom by (x-1) ?

why not multiply top and bottom of 2x/1 by (x-1) ?

you might be mixing up + and =

Show me what that step would look like please.

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I'll let you finish

I'm essentially multiplying by 1
(x-1)/(x-1) = 1

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