Question

@mathmate

Answer 1

@mathmate

Answer 2

@Nnesha its not spam ;)

Answer 3

@Nnesha its not spam ;)

Answer 4

Yes!

Answer 5

What's your question?

Answer 6

O_o question? XD 2+2=?

Answer 7

Not allowed to give direct answers. Check the drawing:

Answer 8

4 LOL

Answer 9

@mathmate ;-;

Answer 10

That means:
1. i dont like this and
2. i dont understand this OR
3. i dont want to do this
Which one/ones?

Answer 11

3 +_+

Answer 12

gimme a minute!

Answer 13

Calculate \(\large \frac{1}{2}+\frac{2}{3}\). Show work.

Answer 14

7/6

Answer 15

Calculate 12+23. \(\huge \color{red}{Show work.}\)

Answer 16

Why are we doing this? .-.

Answer 17

Because it is the same way you do this as with rational fractions.
This is a pretest!

Answer 18

\[12+23=35\]

Answer 19

idk how to show work for that .-.

Answer 20

2+1=3
3+2=5

Answer 21

ok, that's how it works:
Calculate \(\large \frac{1}{2}+\frac{2}{3}\), show work.

Answer 22

\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}\)

Answer 23

Do you know why we multiply the first fraction by three, and the second by two?

Answer 24

to get a common denominator

Answer 25

Exactly!

Answer 26

Just a review of terms:
1/2 = 2/4 because they are __________________

Answer 27

equivalent

Answer 28

exactly "equivalent fractions"! That's a good start! You know your stuff.

Answer 29

:)

Answer 30

The idea of adding and subtracting rational fraction is exactly the same!

Answer 31

We multiply top and bottom of each term by a factor so that the denominator becomes the common denominator.

Answer 32

Since each term remains an equivalent fraction, the answer will not change, BUT...

Answer 33

we now only have to add the numerators of the equivalent fractions, just like the numerical example.

Answer 34

\(\large \frac{1}{2}+\frac{2}{3}=\frac{1*3}{2*3}+\frac{2*2}{3*2}=\frac{3}{6}+\frac{4}{6}=\frac{7}{6}\)

Answer 35

Sorry it took a while!

Answer 36

this looks easy :)

Answer 37

The next one will be equally easy!
gimme a minute.

Answer 38

Calculate \(\large \frac{1}{x}+\frac{2}{x^2}\) show work!

Answer 39

\[\huge~\frac{ 1*x }{ x*x }+\frac{ 2 }{ x^2}=\frac{ 1x+2 }{ x^2 }\]

Answer 40

Excellent! Just remember that 1x is usually written as x, except when showing work.

Answer 41

Now:
Calculate \(\large \frac{2}{(x-1)}-\frac{1}{(x-1)^2}\) show work!

Answer 42

\[\frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }=\frac{ 1(x-1) }{ (x-1) }=1\]

Answer 43

The denominator is still meant to be (x-1)^2 ?
You need to do the math on the numerators. Give it another shot!

Answer 44

T_T not allowed!

Answer 45

;-;

Answer 46

i dont get it

Answer 47

Good! I'll explain!

Answer 48

\(\large \frac{ 2*(x-1) }{ (x-1)*(x-1) }-\frac{ 1 }{ (x-1)^2 }\)
is perfect!

Answer 49

At this point, both denominators are identical, so we can do the math just on the numerators
\(\large \frac{ 2*(x-1) -1 }{ (x-1)^2 }\)

Answer 50

following so far?

Answer 51

yes

Answer 52

Now can you do the math (on the numerator) and finish?

Answer 53

\[\frac{ x(x-1) }{ (x-1)^2 }\]

Answer 54

You need to distribute the first term, and add like terms,
2*(x-1)-1 = 2x-2 -1 =2x-3

Answer 55

following?

Answer 56

Yes

Answer 57

So the final answer is....

Answer 58

\[\huge\frac{ x(x-1) }{ 2x-3 }\]

Answer 59

>_<

Answer 60

wrong spot...

Answer 61

\[\frac{ 2x-3 }{ (x-1)^2 }\]

Answer 62

Whew!
Yes, excellent!

Answer 63

One thing we did without saying it is...nothing.
We did "nothing" to factorize the numerator, because it was "obvious" there are no factors.

Answer 64

In general, after the arithmetic on the numerator, we need to factor the numerator, can you tell me why?

Answer 65

:/ so its easier to multiply or is it to break down the numbers? :/

Answer 66

So that we can cancel (with condition) IF there are common factors between the numerator and denominator, just like what we did before in the simplification.
Does that make sense?

Answer 67

yes

Answer 68

Are there any points to clarify, or are we good?

Answer 69

We're good.

Answer 70

ok, now, try this:
Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\) show work!

Answer 71

ummm O_O

Answer 72

Look at the denominators.

Answer 73

there are no common factors!
So you just "cross multiply", which is a short cut when there are no common factors.
like
\(\large \frac{1}{2}+\frac{3}{5}= \frac{1*5}{2*5}+\frac{2*3}{x*5}=\frac{1*5+2*3}{2*5}=\frac{11}{10}\)

Answer 74

* 2*5 in denominator

Answer 75

O_o

Answer 76

i thought u can only do that if there is an equal sighn :/

Answer 77

It's not a real cross multiplication, but it works in a similar way.

Answer 78

The pattern helps when there are no common factors.
You can use that in the new problem.

Answer 79

Calculate \(\large \frac{3}{x}-\frac{2}{x-1}\), show work!
Remember that when there is no common factors in the denominators, the common denominator is just the product, namely x(x-1)

Answer 80

:?

Answer 81

\(\large \frac{3}{x}-\frac{2}{x-1}=\frac{3(x-1)}{x(x-1)}-\frac{2x}{(x-1)x}=\frac{3(x-1)-2x}{x(x-1)}\)

Answer 82

hmm ok makes a bit more sense

Answer 83

Good!
you can finish it?

Answer 84

\[\frac{ x-1 }{ x(x-1) }\]