18jonea
  • 18jonea
Add/Subtract: : Simplify and state the domain.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
18jonea
  • 18jonea
|dw:1440004351229:dw|
18jonea
  • 18jonea
@Michele_Laino
freckles
  • 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.

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18jonea
  • 18jonea
the plus is suppose to be over before the 1/4
freckles
  • freckles
\[\frac{7}{x+6}-\frac{x+2}{2x+2}+\frac{1}{4}?\]
18jonea
  • 18jonea
yes
freckles
  • 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
  • 18jonea
no?
freckles
  • freckles
what does that mean exactly
freckles
  • freckles
"no?" I don't understand the response to what I said.
freckles
  • freckles
does that mean you don't know how to find a common denominator ?
18jonea
  • 18jonea
i dont know how to find the common denominator
freckles
  • freckles
oh okay
freckles
  • freckles
well if you notice the denominators are x+6 and 2(x+1) and 4
freckles
  • freckles
so a common denominator could be (x+6)*(x+1)*4
freckles
  • freckles
the first fraction... \[\frac{7}{x+6}\] how can we write this as something over/[4(x+6)(x+1)]?
18jonea
  • 18jonea
{4(x+6)(x+1)}
freckles
  • 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
  • freckles
now look at the second fraction
freckles
  • 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
  • 18jonea
x+6
freckles
  • freckles
and also 2 right since 2*2=4
18jonea
  • 18jonea
i am so confused
freckles
  • 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
  • freckles
we wanted 4(x+1)(x+6) on bottom
freckles
  • freckles
we had 2(x+1) on bottom
freckles
  • freckles
2(x+1)*what=4(x+1)(x+6)
freckles
  • 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
  • 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
  • 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
  • freckles
you still need to do the same thing you did to the first two fractions to the third
freckles
  • freckles
what do we need to multiply 4 by so that we have 4(x+1)(x+6)
18jonea
  • 18jonea
x+1 x+6
freckles
  • freckles
right so multiply top and bottom by (x+1)(x+6)
freckles
  • 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
  • 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
  • freckles
you should multiply everything out on top and combine any like terms
freckles
  • freckles
also notice the domain can be found from the start
freckles
  • 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
  • 18jonea
why is there a x+2
18jonea
  • 18jonea
and x+1
freckles
  • freckles
the second fraction had x+2 in the numerator
freckles
  • freckles
and the second fraction also had 2x+2 on bottom 2x+2=2(x+1)
18jonea
  • 18jonea
right
18jonea
  • 18jonea
is the domain all real numbers except when x = 6 or 1
18jonea
  • 18jonea
@freckles
18jonea
  • 18jonea
wait it would be -6 right?
freckles
  • freckles
-6 is right 1 is incorrect this is because 1+1=0 isn't a true equation
freckles
  • freckles
-1+1=0 is true
freckles
  • freckles
x+1=0 means x=-1
freckles
  • freckles
the domain is all real numbers except x=-1 or x=-6
18jonea
  • 18jonea
right i just figured that out
freckles
  • freckles
did you ever try multiplying the top and combining like terms yet?
18jonea
  • 18jonea
i am trying to figure out the second equation you did... can you show me how you did it again please
freckles
  • freckles
fraction?
18jonea
  • 18jonea
yes the second fraction
freckles
  • freckles
this was the second fraction \[-\frac{x+2}{2(x+1)}\]
freckles
  • freckles
we wanted the bottom to be 4(x+1)(x+6)
freckles
  • freckles
the bottom is missing factors 2 and (x+6)
freckles
  • freckles
so we multiply top and bottom by 2(x+6)
18jonea
  • 18jonea
so how would it look
freckles
  • 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
  • 18jonea
ok i am goimg to try and multiply now
freckles
  • freckles
ok
18jonea
  • 18jonea
yeah i have no idea how to even start... can you help
18jonea
  • 18jonea
@freckles
18jonea
  • 18jonea
wouldnt the second one be 2x squared +16x + 24
18jonea
  • 18jonea
@freckles
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
  • 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
  • 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
  • freckles
\[\frac{28x+28-2x^2-16x-24+x^2+7x+6}{4(x+1)(x+6)}\]
freckles
  • freckles
combine like terms on top
18jonea
  • 18jonea
to combine all of them i got -x^2 +12x +10 All divided by 4(x+6)(x+1)
18jonea
  • 18jonea
is that right?
freckles
  • 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
  • 18jonea
yes thats what i got
18jonea
  • 18jonea
sorry i meanyt 19 x
18jonea
  • 18jonea
then how would i siplify it
freckles
  • freckles
in my mind \[\frac{-x^2+19x+10}{4(x+1)(x+6)}\] this is simplified
freckles
  • freckles
however your teacher may want you to multiply the bottom out
freckles
  • freckles
not sure in your teacher's mind
18jonea
  • 18jonea
thank you
freckles
  • freckles
np
18jonea
  • 18jonea
can you help me with one more of these problems that similar..
freckles
  • freckles
you can show me your work and I can help you fix it if it needs any fixing
18jonea
  • 18jonea
ok i will
18jonea
  • 18jonea
\[\frac{ 1 }{ 2} +\frac{ 15 }{ 2x-14 } - \frac{ 3x-5 }{ x^2-7x }\]
18jonea
  • 18jonea
first one stays the same 2(x-7) x(x-7)
freckles
  • freckles
check mark so far! :)
18jonea
  • 18jonea
domain is all real numbers except when x=7
freckles
  • freckles
well or x=0
freckles
  • freckles
you have x on the bottom
freckles
  • freckles
on that 3rd fraction
freckles
  • 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
  • 18jonea
ok
freckles
  • 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
  • 18jonea
you tell me and i try to figure out the numerators
freckles
  • 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
  • 18jonea
\[\frac{ 1 }{ 2 } * x(x-7)\]
18jonea
  • 18jonea
x * x(x-7)
18jonea
  • 18jonea
is that right
freckles
  • freckles
well you multiply top and bottom by x(x-7)
freckles
  • freckles
\[\frac{1}{2} \cdot \frac{x(x-7)}{x(x-7)}\]
freckles
  • freckles
\[\frac{x(x-7)}{2x(x-7)}\] but yep this right for the first fraction
18jonea
  • 18jonea
ok
18jonea
  • 18jonea
the next one you only need to multiply top and bottom by x
18jonea
  • 18jonea
and the last one you only multiply it by 2
freckles
  • 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
  • 18jonea
thats what i got
freckles
  • freckles
so now we have the same denominators we can now combine the fractions
freckles
  • 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
  • freckles
multiply top out and combine like terms
18jonea
  • 18jonea
x^2 +2x - 10
18jonea
  • 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
  • 18jonea
@freckles
freckles
  • freckles
sorry my internet stopped working for a few minutes there
freckles
  • freckles
and you can't cancel out common terms but you can cancel out common factors
freckles
  • 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
  • freckles
@18jonea
18jonea
  • 18jonea
so what would be the final equation
18jonea
  • 18jonea
@freckles
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
  • 18jonea
how did you get 14x and -20
18jonea
  • 18jonea
the last fraction you are supppose to subtract
freckles
  • freckles
the problem I have is: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)}\]
freckles
  • freckles
is that not the right problem?
18jonea
  • 18jonea
sorry it was subtract the last fraction
18jonea
  • 18jonea
thats how i got \[\frac{ x^2 +2x-10 }{ 2x(x-7) }\]
freckles
  • 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
  • 18jonea
wait it is -10
freckles
  • freckles
-2(-5)=10
18jonea
  • 18jonea
why is it -2 for the rest it was +2
freckles
  • 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
  • 18jonea
why is it -2 cause it has +2 at bottom
18jonea
  • 18jonea
i have 6x-10
18jonea
  • 18jonea
hence how i got x2+2x−10 / 2x(x−7)
freckles
  • 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
  • freckles
do you see that red subtraction symbol I just brought it down
18jonea
  • 18jonea
got it thank you!

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