vera_ewing
  • vera_ewing
How do you solve this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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vera_ewing
  • vera_ewing
jim_thompson5910
  • jim_thompson5910
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
vera_ewing
  • vera_ewing
Wait why do I subtract 1 from both sides?!

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jim_thompson5910
  • jim_thompson5910
because they want to know what \(\large \boxed{x}-1\) is equivalent to
vera_ewing
  • vera_ewing
So we just put a -1 on both sides? That makes no sense...
jim_thompson5910
  • jim_thompson5910
you start off with \(\large \boxed{x}\) and do that operation to get \(\large \boxed{x}-1\)
vera_ewing
  • vera_ewing
But there is no -1 originally. Where did it come from?
jim_thompson5910
  • jim_thompson5910
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
jim_thompson5910
  • jim_thompson5910
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
vera_ewing
  • vera_ewing
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....
jim_thompson5910
  • jim_thompson5910
This is what the full step by step picture looks like \[\Large \boxed{x} = \frac{x+3}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1*\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3}{x-1}-\frac{x-1}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-(x-1)}{x-1}\] \[\Large \boxed{x}-1 = \frac{x+3-x+1}{x-1}\] \[\Large \boxed{x}-1 = \frac{4}{x-1}\]
jim_thompson5910
  • jim_thompson5910
I'm using the subtraction property of equality if a = b, then a-c = b-c
vera_ewing
  • vera_ewing
Ohh okay. And why is there a box around the x on the left?
jim_thompson5910
  • jim_thompson5910
because that's how they set up the notation in the problem
vera_ewing
  • vera_ewing
Does it mean anything?
jim_thompson5910
  • jim_thompson5910
it's just their fancy way of defining a function
jim_thompson5910
  • jim_thompson5910
they could have easily said "let y = ..." or "let f(x) = ..."
jim_thompson5910
  • jim_thompson5910
why they chose a box, who knows
vera_ewing
  • vera_ewing
Okay. Can you please give me a similar problem to solve?
vera_ewing
  • vera_ewing
I need to be able to do these on my own.
jim_thompson5910
  • jim_thompson5910
ok what is \[\Large \boxed{x} +2x\] equivalent to? (same function from the previous problem)
vera_ewing
  • vera_ewing
I'm not sure :(
jim_thompson5910
  • jim_thompson5910
replace the box x with what it is equal to then simplify
vera_ewing
  • vera_ewing
4/x-1 ?
jim_thompson5910
  • jim_thompson5910
no
jim_thompson5910
  • jim_thompson5910
box x was equal to (x+3)/(x-1)
jim_thompson5910
  • jim_thompson5910
so saying \[\Large \boxed{x}+2x\]is the same as saying (x+3)/(x-1) + 2x. Simplify from here
vera_ewing
  • vera_ewing
What would the first step be?
jim_thompson5910
  • jim_thompson5910
you need to get the denominators the same
jim_thompson5910
  • jim_thompson5910
2x = 2x/1 has a denominator of 1
jim_thompson5910
  • jim_thompson5910
how can you transform that 1 into (x-1)
vera_ewing
  • vera_ewing
So (x+3)/(x-1) + 2x/1 ?
jim_thompson5910
  • jim_thompson5910
how can you get 2x/1 to have a denominator of x-1
vera_ewing
  • vera_ewing
Multiply 1 on both sides?
jim_thompson5910
  • jim_thompson5910
why not multiply top and bottom by (x-1) ?
jim_thompson5910
  • jim_thompson5910
why not multiply top and bottom of 2x/1 by (x-1) ?
jim_thompson5910
  • jim_thompson5910
you might be mixing up + and =
vera_ewing
  • vera_ewing
Show me what that step would look like please.
jim_thompson5910
  • jim_thompson5910
|dw:1434499487201:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1434499531547:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1434499550135:dw|
jim_thompson5910
  • jim_thompson5910
|dw:1434499580374:dw|
jim_thompson5910
  • jim_thompson5910
I'll let you finish
vera_ewing
  • vera_ewing
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.
jim_thompson5910
  • jim_thompson5910
I'm essentially multiplying by 1 (x-1)/(x-1) = 1
jim_thompson5910
  • jim_thompson5910
|dw:1434499775976:dw|