Add/Subtract: : Simplify and state the domain.

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Add/Subtract: : Simplify and state the domain.

Mathematics
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|dw:1440004351229:dw|
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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the plus is suppose to be over before the 1/4
\[\frac{7}{x+6}-\frac{x+2}{2x+2}+\frac{1}{4}?\]
yes
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
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)}
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 ... \[\frac{7}{x+6} \\ \text{ multiply top and bottom by } 4(x+1) \\ \frac{7}{x+6} \cdot \frac{4(x+1)}{4(x+1)}=\frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}\]
now look at the second fraction
\[-\frac{x+2}{2(x+1)}\] remember we want 4(x+6)(x+1) on bottom so what is this bottom lacking in order to get that?
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)
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)
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)
so far we have: \[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\]
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)
\[\frac{1}{4}=\frac{1(x+1)(x+6)}{4(x+1)(x+6)} \text{ or just } =\frac{(x+1)(x+6)}{4(x+1)(x+6)}\]
so this gives us: \[\frac{7}{x+6}-\frac{x+2}{2(x+1)}+\frac{1}{4} \\ = \\ \frac{7 \cdot 4(x+1)}{4(x+6)(x+1)}-\frac{2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4} \\ \text{ so we are already able to combine the first two fractions } \\ \frac{7 \cdot 4(x+1)-2(x+6)(x+2)}{4(x+1)(x+6)}+\frac{1}{4}\] \[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)}\]
you should multiply everything out on top and combine any like terms
also notice the domain can be found from the start
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)
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
\[-\frac{x+2}{2(x+1)} \cdot \frac{2(x+6)}{2(x+6)} \\ =-\frac{2(x+2)(x+6)}{2 \cdot 2(x+1)(x+6)} \\ = -\frac{2(x+2)(x+6)}{4(x+1)(x+6)}\]
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
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
\[\frac{7 \cdot 4(x+1)-2(x+6)(x+2)+(x+1)(x+6)}{4(x+1)(x+6)} \\ \frac{28(x+1)-2x^2-16x-24+(x+1)(x+6)}{4(x+1)(x+6)}\]
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?
\[28x+28-2x^2-16x-24+x^2+7x+6 \\ \text{ reorder terms } \\ (-2x^2+x^2)+(28x-16x+7x)+(28-24+6) \\ (-2+1)x^2+(28-16+7)x+(28-24+6) \\ -1x^2+19x+10 \\ -x^2+19x+10\] is what I got for the numerator you might want to check my work
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
\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ \text{ so we want } x-7 \neq 0 \text{ and also we want } x \neq 0\]
ok
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?
you tell me and i try to figure out the numerators
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)
\[\frac{ 1 }{ 2 } * x(x-7)\]
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
(well 2/2 but I get what you are saying) right!!! so we have: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)}\]
thats what i got
so now we have the same denominators we can now combine the fractions
\[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)}\]
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 you should have: \[\frac{1}{2}+\frac{15}{2(x-7)}+\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}+\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x+2(3x-5)}{2x(x-7)} \\ =\frac{x^2+14x-20}{2x(x-7)}\]
so what would be the final equation
\[\frac{x^2+14x-20}{2x(x-7)}\] and you can multiply to the bottom out if you want or if your teacher wants but to me this is simplified
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 \[\frac{ x^2 +2x-10 }{ 2x(x-7) }\]
\[\frac{1}{2}+\frac{15}{2(x-7)}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)}\] \[=\frac{x^2+2x+10}{2x(x-7)}\]
wait it is -10
-2(-5)=10
why is it -2 for the rest it was +2
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
why is it -2 cause it has +2 at bottom
i have 6x-10
hence how i got x2+2x−10 / 2x(x−7)
\[\frac{1}{2}+\frac{15}{2(x-7)}\color{red}-\frac{3x-5}{x(x-7)} \\ =\frac{x(x-7)}{2x(x-7)}+\frac{15x}{2x(x-7)}\color{red}-\frac{2(3x-5)}{2x(x-7)} \\ =\frac{x(x-7)+15x\color{red}-2(3x-5)}{2x(x-7)} \\ =\frac{x^2-7x+15x-6x+10}{2x(x-7)} \\ =\frac{x^2+2x+10}{2x(x-7)}\]
do you see that red subtraction symbol I just brought it down
got it thank you!

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