Add/Subtract: : Simplify and state the domain.

- 18jonea

Add/Subtract: : Simplify and state the domain.

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 18jonea

|dw:1440004351229:dw|

- 18jonea

@Michele_Laino

- freckles

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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 18jonea

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

- freckles

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

- 18jonea

yes

- freckles

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

- 18jonea

no?

- freckles

what does that mean exactly

- freckles

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

- freckles

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

- 18jonea

i dont know how to find the common denominator

- freckles

oh okay

- freckles

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

- freckles

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

- freckles

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

- 18jonea

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

- freckles

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 ...
\[\frac{7}{x+6} \\ \text{ multiply top and bottom by } 4(x+1) \\ \frac{7}{x+6} \cdot \frac{4(x+1)}{4(x+1)}=\frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}\]

- freckles

now look at the second fraction

- freckles

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

- 18jonea

x+6

- freckles

and also 2 right since 2*2=4

- 18jonea

i am so confused

- freckles

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

- freckles

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

- freckles

we had 2(x+1) on bottom

- freckles

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

- freckles

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)

- freckles

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)

- freckles

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

- freckles

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

- freckles

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

- 18jonea

x+1 x+6

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- freckles

also notice the domain can be found from the start

- freckles

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)

- 18jonea

why is there a x+2

- 18jonea

and x+1

- freckles

the second fraction had x+2 in the numerator

- freckles

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

- 18jonea

right

- 18jonea

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

- 18jonea

@freckles

- 18jonea

wait it would be -6 right?

- freckles

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

- freckles

-1+1=0 is true

- freckles

x+1=0 means x=-1

- freckles

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

- 18jonea

right i just figured that out

- freckles

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

- 18jonea

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

- freckles

fraction?

- 18jonea

yes the second fraction

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- 18jonea

so how would it look

- freckles

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

- 18jonea

ok i am goimg to try and multiply now

- freckles

ok

- 18jonea

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

- 18jonea

@freckles

- 18jonea

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

- 18jonea

@freckles

- freckles

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

- freckles

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

- freckles

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

- freckles

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

- freckles

combine like terms on top

- 18jonea

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

- 18jonea

is that right?

- freckles

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

- 18jonea

yes thats what i got

- 18jonea

sorry i meanyt 19 x

- 18jonea

then how would i siplify it

- freckles

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

- freckles

however your teacher may want you to multiply the bottom out

- freckles

not sure in your teacher's mind

- 18jonea

thank you

- freckles

np

- 18jonea

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

- freckles

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

- 18jonea

ok i will

- 18jonea

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

- 18jonea

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

- freckles

check mark so far! :)

- 18jonea

domain is all real numbers except when x=7

- freckles

well or x=0

- freckles

you have x on the bottom

- freckles

on that 3rd fraction

- freckles

\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ \text{ so we want } x-7 \neq 0 \text{ and also we want } x \neq 0\]

- 18jonea

ok

- freckles

do you want to try to figure out what the denominators should be so we can combine the fractions? or do you want me to tell you and you try to figure out the new numerators?

- 18jonea

you tell me and i try to figure out the numerators

- freckles

ok notice the denominators are:
2
2(x-7)
x(x-7)
----
if we look at just 2 and 2(x-7) then the denominator would be 2(x-7)
but we also have x(x-7) for the third bottom
so looking at 2(x-7) and x(x-7) the new denominator of all them should be
2x(x-7)
this is because all the denominators given go into 2x(x-7)
that I can think of something to multiply all of your original denominators
to get this new denominator of 2x(x-7)
for example:
2*x(x-7) is 2x(x-7)
and
2(x-7)*x is 2x(x-7)
and
x(x-7)*2 is 2x(x-7)

- 18jonea

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

- 18jonea

x * x(x-7)

- 18jonea

is that right

- freckles

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

- freckles

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

- freckles

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

- 18jonea

ok

- 18jonea

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

- 18jonea

and the last one you only multiply it by 2

- freckles

(well 2/2 but I get what you are saying)
right!!!
so we have:
\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)}\]

- 18jonea

thats what i got

- freckles

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

- freckles

\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)}\]

- freckles

multiply top out
and combine like terms

- 18jonea

x^2 +2x - 10

- 18jonea

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

- 18jonea

@freckles

- freckles

sorry my internet stopped working for a few minutes there

- freckles

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

- freckles

so you should have:
\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)} \\ =\frac{x^2+14x-20}{2x(x-7)}\]

- freckles

@18jonea

- 18jonea

so what would be the final equation

- 18jonea

@freckles

- freckles

\[\frac{x^2+14x-20}{2x(x-7)}\]
and you can multiply to the bottom out if you want
or if your teacher wants
but to me this is simplified

- 18jonea

how did you get 14x and -20

- 18jonea

the last fraction you are supppose to subtract

- freckles

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

- freckles

is that not the right problem?

- 18jonea

sorry it was subtract the last fraction

- 18jonea

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

- freckles

\[\frac{1}{2}+\frac{15}{2(x-7)}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)}\]
\[=\frac{x^2+2x+10}{2x(x-7)}\]

- 18jonea

wait it is -10

- freckles

-2(-5)=10

- 18jonea

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

- freckles

you have -2(3x-5)
distributive property says a(b+c)=ab+ac
so this means -2(3x-5)
is -2(3x)+(-2)(-5)
or -6x+10

- 18jonea

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

- 18jonea

i have 6x-10

- 18jonea

hence how i got x2+2x−10 / 2x(x−7)

- freckles

\[\frac{1}{2}+\frac{15}{2(x-7)}\color{red}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}\color{red}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x\color{red}-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)} \\ =\frac{x^2+2x+10}{2x(x-7)}\]

- freckles

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

- 18jonea

got it thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.