What is the solution to the rational equation x over x squared minus 9 plus 1 over x plus 3 equals 1 over 4x minus 12 ?

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What is the solution to the rational equation x over x squared minus 9 plus 1 over x plus 3 equals 1 over 4x minus 12 ?

Mathematics
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\[\huge\rm \frac{ x }{ x^2-9}+\frac{ 1 }{ x+3}=\frac{ 1 }{ 4x-12 }\] like this ?
yes
alright first solve left side apply difference of square method to factor x^2-9 \[\rm a^2-b^2=(a+b)(a-b)\]

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take square of both terms write ur answer as (sqrt of 1st term + sqrt of 2nd term) (sqrt of 1st term - sqrt of 2nd term) here is an example \[\rm x^2-16=(x+4)(x-4)\]
x = -9 over 7 x = 15 x = 15 over 7 x = 9
these are the answers i have
ok. so what are the factors of x^2-9 ?
(x-9)(x+9)
hmm not what is the square root of 9 ?
3
|dw:1440351548850:dw|
yes right so it should be (x-3)(x+3)
\[\huge\rm \frac{ x }{ (x-3)(x+3)}+\frac{ 1 }{ x+3}=\frac{ 1 }{ 4x-12 }\] now find common denominator what is the common denominator at left side ?
so 4(3)-12?
no.. common denominator\[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] just look at red part what is common ?
3
its not just 3 (x-3) is one term so x+3 is common
.. \[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] \[\huge\rm \frac{ x +1(x-3)}{ (x-3)(x+3) }\] when you find common denominator you should multiply `numerator` of `1st` fraction by the `denominator` of `2nd` fraction and multiply numerator of `2nd` fraction by the denomiantor of 1st fraction
\[-\frac{ 9 }{ 7 }\]
combine like terms .. \[\huge\rm \frac{ x +1x-3}{ (x-3)(x+3) }\] x+1x-3= ??
? is this right
what ??
no that's not right
what are `like terms` in this equation x+x-3 ?
x
yes right so combine them x+x = ?
x^2?
no when you multiply same bases then you should add their exponents no just add the coefficient 1x+1x=(1+1)x = ?
im confused
okay so here we need to `combine` x+x which is same as 1x+1x one is invisible we here is an example 3y+5y = (3+5)y = 8y just add the coefficient \[\rm x \times x=x^2\] when you multiply them `then` you should add their exponents
|dw:1440353412609:dw| leading coefficient a number front of the variable
1x+1x = ?
1x
add their coefficient \[1x+1x=(1+1)x= ??x\]
2x
yes right .. \[\huge\rm \frac{ x }{ \color{Red}{(x-3)(x+3)}}+\frac{ 1 }{\color{Red}{ x+3}}=\frac{ 1 }{ 4x-12 }\] \[\huge\rm \frac{ 2x-3}{ (x-3)(x+3) }=\frac{1}{4x-12}\] now foil this (x-3)(x+3) are you familiar with the foil method ?
x = 9 is this right?
no... show ur work so i'll look ovr to find mistakes :=)
idk i dont understand
so you don't know how to foil two parentheses ?
ido x^2 +3x-3x-9?
yes right which is same as x^2-9 \[\frac{ 2x-3 }{ x^2-9 }=\frac{ 1 }{ 4x+12 }\]now cross multiply
|dw:1440355272983:dw|

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