How do you solve this?

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How do you solve this?

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They provided the definition of \(\Large \boxed{x}\) \[\Large \boxed{x} = \frac{x+3}{x-1}\] subtract 1 from both sides to get \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\] now simplify the right hand side
Wait why do I subtract 1 from both sides?!

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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?
maybe it might make more sense if y = 2x was given to you if that's the case, then you would subtract 1 from both sides to get y-1 = 2x-1
I'm doing that operation of subtract 1 from both sides to turn \(\large \boxed{x}\) into \(\large \boxed{x}-1\) basically it's like turning y into y-1
I'm so sorry, but why are we allowed to subtract 1 from both sides? There is no 1 on either side of the equation....
This is what the full step by step picture looks like \[\Large \boxed{x} = \frac{x+3}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1*\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-(x-1)}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-x+1}{x-1}\] \[\Large \boxed{x}-1 = \frac{4}{x-1}\]
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 ?
no
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
Hold on, this all makes so much sense, but I just don't understand what property lets us multiply (x-1) on the top and bottom.
I'm essentially multiplying by 1 (x-1)/(x-1) = 1
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