At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions.

See more answers at brainly.com

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this and **thousands** of other questions

|dw:1440004351229:dw|

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

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

yes

no?

what does that mean exactly

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

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

i dont know how to find the common denominator

oh okay

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

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

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

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

now look at the second fraction

x+6

and also 2 right since 2*2=4

i am so confused

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

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

we had 2(x+1) on bottom

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

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

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

x+1 x+6

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

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

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

also notice the domain can be found from the start

why is there a x+2

and x+1

the second fraction had x+2 in the numerator

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

right

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

wait it would be -6 right?

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

-1+1=0 is true

x+1=0 means x=-1

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

right i just figured that out

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

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

fraction?

yes the second fraction

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

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

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

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

so how would it look

ok i am goimg to try and multiply now

ok

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

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

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

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

combine like terms on top

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

is that right?

yes thats what i got

sorry i meanyt 19 x

then how would i siplify it

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

however your teacher may want you to multiply the bottom out

not sure in your teacher's mind

thank you

np

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

you can show me your work and I can help you fix it if it needs any fixing

ok i will

\[\frac{ 1 }{ 2} +\frac{ 15 }{ 2x-14 } - \frac{ 3x-5 }{ x^2-7x }\]

first one stays the same
2(x-7)
x(x-7)

check mark so far! :)

domain is all real numbers except when x=7

well or x=0

you have x on the bottom

on that 3rd fraction

ok

you tell me and i try to figure out the numerators

\[\frac{ 1 }{ 2 } * x(x-7)\]

x * x(x-7)

is that right

well you multiply top and bottom by x(x-7)

\[\frac{1}{2} \cdot \frac{x(x-7)}{x(x-7)}\]

\[\frac{x(x-7)}{2x(x-7)}\]
but yep this right for the first fraction

ok

the next one you only need to multiply top and bottom by x

and the last one you only multiply it by 2

thats what i got

so now we have the same denominators we can now combine the fractions

multiply top out
and combine like terms

x^2 +2x - 10

so i can take the 2x out of the top and bottom and then i would get
\[\frac{ x^2-10 }{ x-7 }\]

sorry my internet stopped working for a few minutes there

and you can't cancel out common terms but you can cancel out common factors

so what would be the final equation

how did you get 14x and -20

the last fraction you are supppose to subtract

the problem I have is:
\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)}\]

is that not the right problem?

sorry it was subtract the last fraction

thats how i got \[\frac{ x^2 +2x-10 }{ 2x(x-7) }\]

wait it is -10

-2(-5)=10

why is it -2
for the rest it was +2

why is it -2 cause it has +2 at bottom

i have 6x-10

hence how i got x2+2x−10 / 2x(x−7)

do you see that red subtraction symbol
I just brought it down

got it thank you!