Question

Add/Subtract: : Simplify and state the domain.

Answer 1

@Michele_Laino

Answer 2

can't really understand this? what is going on with that 1/4 over there...
And I see you have - + in between the first two fractions.

Answer 3

the plus is suppose to be over before the 1/4

Answer 4

\[\frac{7}{x+6}-\frac{x+2}{2x+2}+\frac{1}{4}?\]

Answer 5

yes

Answer 6

well first 2x+2 can be written as 2(x+1)
\[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4}\]

now look at the bottoms of all the fractions and ask yourself how you can find a common denominator

Answer 7

no?

Answer 8

what does that mean exactly

Answer 9

"no?"
I don't understand the response to what I said.

Answer 10

does that mean you don't know how to find a common denominator ?

Answer 11

i dont know how to find the common denominator

Answer 12

oh okay

Answer 13

well if you notice the denominators are
x+6
and
2(x+1)
and
4

Answer 14

so a common denominator could be (x+6)*(x+1)*4

Answer 15

the first fraction...
\[\frac{7}{x+6}\]
how can we write this as something over/[4(x+6)(x+1)]?

Answer 16

{4(x+6)(x+1)}

Answer 17

notice on bottom it is lacking the 4 and the (x+1)
so just multiply both top and bottom by 4(x+1)
that is how you can get this new bottom we speak of ...
\[\frac{7}{x+6} \\ \text{ multiply top and bottom by } 4(x+1) \\ \frac{7}{x+6} \cdot \frac{4(x+1)}{4(x+1)}=\frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}\]

Answer 18

now look at the second fraction

Answer 19

\[-\frac{x+2}{2(x+1)}\]
remember we want 4(x+6)(x+1) on bottom
so what is this bottom lacking in order to get that?

Answer 20

x+6

Answer 21

and also 2 right since 2*2=4

Answer 22

i am so confused

Answer 23

\[-\frac{x+2}{2(x+1)} \cdot \frac{2(x+6)}{2(x+6)} =-\frac{2(x+2)(x+6)}{2 \cdot 2 (x+6)(x+1)}\]

Answer 24

we wanted 4(x+1)(x+6) on bottom

Answer 25

we had 2(x+1) on bottom

Answer 26

2(x+1)*what=4(x+1)(x+6)

Answer 27

divide both sides 2(x+1)
\[what=\frac{4(x+1)(x+6)}{2(x+1)}=\frac{4}{2} \frac{(x+1)}{(x+1)} \frac{x+6}{1} \\ what=2 (1)(x+6) \\ what=2(x+6)\]

so
2(x+1)*2(x+6)=4(x+1)(x+6)

Answer 28

so what we needed to do to get the bottom 4(x+1)(x+6) for the second fraction was multiply the second fraction on top and bottom by 2(x+6)

Answer 29

so far we have:
\[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\]

Answer 30

you still need to do the same thing you did to the first two fractions to the third

Answer 31

what do we need to multiply 4 by so that we have 4(x+1)(x+6)

Answer 32

x+1 x+6

Answer 33

right so multiply top and bottom by (x+1)(x+6)

Answer 34

\[\frac{1}{4}=\frac{1(x+1)(x+6)}{4(x+1)(x+6)} \text{ or just } =\frac{(x+1)(x+6)}{4(x+1)(x+6)}\]

Answer 35

so this gives us:
\[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\]
\[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)}\]

Answer 36

you should multiply everything out on top and combine any like terms

Answer 37

also notice the domain can be found from the start

Answer 38

you don't want any of the fractions to be 0 on bottom
so to find the domain say all real numbers except x+1=0 or x+6=0

(I will let you solve those equations)

Answer 39

why is there a x+2

Answer 40

and x+1

Answer 41

the second fraction had x+2 in the numerator

Answer 42

and the second fraction also had 2x+2 on bottom
2x+2=2(x+1)

Answer 43

right

Answer 44

is the domain all real numbers except when x = 6 or 1

Answer 45

@freckles

Answer 46

wait it would be -6 right?

Answer 47

-6 is right
1 is incorrect this is because 1+1=0 isn't a true equation

Answer 48

-1+1=0 is true

Answer 49

x+1=0 means x=-1

Answer 50

the domain is all real numbers except x=-1 or x=-6

Answer 51

right i just figured that out

Answer 52

did you ever try multiplying the top and combining like terms yet?

Answer 53

i am trying to figure out the second equation you did... can you show me how you did it again please

Answer 54

fraction?

Answer 55

yes the second fraction

Answer 56

this was the second fraction
\[-\frac{x+2}{2(x+1)}\]

Answer 57

we wanted the bottom to be 4(x+1)(x+6)

Answer 58

the bottom is missing factors 2 and (x+6)

Answer 59

so we multiply top and bottom by 2(x+6)

Answer 60

so how would it look

Answer 61

\[-\frac{x+2}{2(x+1)} \cdot \frac{2(x+6)}{2(x+6)} \\ =-\frac{2(x+2)(x+6)}{2 \cdot 2(x+1)(x+6)} \\ = -\frac{2(x+2)(x+6)}{4(x+1)(x+6)}\]

Answer 62

ok i am goimg to try and multiply now

Answer 63

ok

Answer 64

yeah i have no idea how to even start... can you help

Answer 65

@freckles

Answer 66

wouldnt the second one be 2x squared +16x + 24

Answer 67

@freckles

Answer 68

2(x+2)(x+6)
=
2(x^2+6x+2x+12)
2(x^2+8x+12)
2x^2+16x+24
is right
and then there is a negative in front of that mess
-2(x+2)(x+6)
so it is actually
-2x^2-16x-24

Answer 69

\[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)} \\ \frac{28(x+1)-2x^2-16x-24+(x+1)(x+6)}{4(x+1)(x+6)}\]

Answer 70

28(x+1)=28x+28
and
(x+1)(x+6)=x^2+6x+1x+6
(x+1)(x+6)+x^2+7x+6

Answer 71

\[\frac{28x+28-2x^2-16x-24+x^2+7x+6}{4(x+1)(x+6)}\]

Answer 72

combine like terms on top

Answer 73

to combine all of them i got -x^2 +12x +10 All divided by 4(x+6)(x+1)

Answer 74

is that right?

Answer 75

\[28x+28-2x^2-16x-24+x^2+7x+6 \\ \text{ reorder terms } \\ (-2x^2+x^2)+(28x-16x+7x)+(28-24+6) \\ (-2+1)x^2+(28-16+7)x+(28-24+6) \\ -1x^2+19x+10 \\ -x^2+19x+10\]
is what I got for the numerator
you might want to check my work

Answer 76

yes thats what i got

Answer 77

sorry i meanyt 19 x

Answer 78

then how would i siplify it

Answer 79

in my mind
\[\frac{-x^2+19x+10}{4(x+1)(x+6)}\]
this is simplified

Answer 80

however your teacher may want you to multiply the bottom out

Answer 81

not sure in your teacher's mind

Answer 82

thank you

Answer 83

np

Answer 84

can you help me with one more of these problems that similar..