Question

How do you solve this?

Answer 1

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

Answer 2

Wait why do I subtract 1 from both sides?!

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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....

Answer 10

This is what the full step by step picture looks like
\[\Large \boxed{x} = \frac{x+3}{x-1}\]
\[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1\]
\[\Large \boxed{x}-1 = \frac{x+3}{x-1}-1*\frac{x-1}{x-1}\]
\[\Large \boxed{x}-1 = \frac{x+3}{x-1}-\frac{x-1}{x-1}\]
\[\Large \boxed{x}-1 = \frac{x+3-(x-1)}{x-1}\]
\[\Large \boxed{x}-1 = \frac{x+3-x+1}{x-1}\]
\[\Large \boxed{x}-1 = \frac{4}{x-1}\]

Answer 11

I'm using the subtraction property of equality

if a = b, then a-c = b-c

Answer 12

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

Answer 13

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

Answer 14

Does it mean anything?

Answer 15

it's just their fancy way of defining a function

Answer 16

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

Answer 17

why they chose a box, who knows

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

I'm not sure :(

Answer 22

replace the box x with what it is equal to

then simplify

Answer 23

4/x-1 ?

Answer 24

no

Answer 25

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

Answer 26

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

Answer 27

What would the first step be?

Answer 28

you need to get the denominators the same

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

Multiply 1 on both sides?

Answer 34

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

Answer 35

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

Answer 36

you might be mixing up + and =

Answer 37

Show me what that step would look like please.