How do you solve this?

- vera_ewing

How do you solve this?

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- schrodinger

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- vera_ewing

##### 1 Attachment

- 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

Wait why do I subtract 1 from both sides?!

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## More answers

- jim_thompson5910

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

- vera_ewing

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

- jim_thompson5910

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

- vera_ewing

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

- 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

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

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

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

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

- vera_ewing

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

- jim_thompson5910

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

- vera_ewing

Does it mean anything?

- jim_thompson5910

it's just their fancy way of defining a function

- jim_thompson5910

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

- jim_thompson5910

why they chose a box, who knows

- vera_ewing

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

- vera_ewing

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

- jim_thompson5910

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

- vera_ewing

I'm not sure :(

- jim_thompson5910

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

- vera_ewing

4/x-1 ?

- jim_thompson5910

no

- jim_thompson5910

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

- jim_thompson5910

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

- vera_ewing

What would the first step be?

- jim_thompson5910

you need to get the denominators the same

- jim_thompson5910

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

- jim_thompson5910

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

- vera_ewing

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

- jim_thompson5910

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

- vera_ewing

Multiply 1 on both sides?

- jim_thompson5910

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

- jim_thompson5910

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

- jim_thompson5910

you might be mixing up + and =

- vera_ewing

Show me what that step would look like please.

- jim_thompson5910

|dw:1434499487201:dw|

- jim_thompson5910

|dw:1434499531547:dw|

- jim_thompson5910

|dw:1434499550135:dw|

- jim_thompson5910

|dw:1434499580374:dw|

- jim_thompson5910

I'll let you finish

- 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

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

- jim_thompson5910

|dw:1434499775976:dw|