Maths Dilemma !!?

Question

I cant seem to solve dis stupid problem

Add the difference of

[ x-4/x-3 and x+7/x+8] to x+8/ x(square) + 5x - 24

I ended up gettin 15x - 25/ (x-3) (x-8)

but my friend got

1/x + 8

Pls Tell me which one is correct and HOW ?

Answer 1

X-4)/(x-3)-(x+7)/(x+8) the LCM will be (x-3)(x+8) and the Nr.will be
(X-4)(X+8)-(x+7)(x-3)=>(x^2+4x-32)-(x^2+4x-21)=-11
So the expression i.e. the difference=11/(x-3)(x+8)
(X+8)/(x^2+5x-24)=(x+8)/(x+8)(x-3) which is =1/(x-3)
Now adding –11/(x-3)(x+8)+1/(x-3) the LCM will be (x-3)(x+8)
So the sum is –11+x+8/(x-3)(x+8) =>(x-3)/(x-3)(x+8)=1/(x+8)
Your friend’s answer seems to be right!

Answer 2

x - 4/x - 3 - x + 7/x + 8

The common denominator is (x - 3)(x + 8)

(x - 4)(x + 8)/(x - 3)(x + 9) - (x + 7)(x - 3)/(x - 3)(x + 8)

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Proof of multiplication on the left side numerator

X - 4
x + 8

x² - 4x
+ 8x - 32

x² + 4x - 32

Insert this resultant product into the left side of the equation

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Proof of multiplication of the right side numerator

x + 7
x - 3

x² + 7x
-3x - 21
- - - - - - - -
x² + 4x - 21

Insert this resultant product into the right side of the equation

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x² + 4x - 32/(x - 3)(x + 8) - x² + 4x - 21/(x -3) (x + 8)

x² + 4x - 32 - (x² + 4x - 21)/(x - 3)(x + 8

Combining the left with the right . elimination one of the common denominators. Don't forget the parenthesis in the numerator when combining. The parenthesis following the minus ( - ) sign

x² + 4x - 32 - x² - 4x + 21/(x -3) (x+ 8)

removing parenthesis when a minus (- ) sign is present.
all signs inside the parenthesis change.

- 11/(x - 3)(x + 8)

The x² - x² Cancell

The + 4x - 4x Cancell

-32 + 21 = - 11

The answer is -11/(x -3)(x + 8)

Answer 3

First find the difference
(x-4)/(x-3) - (x+7)/(x+8)
The common denominator is (x-3)(x+8)
(x-4)(x+8) = x^2 + 4x - 28
(x+7)(x-3) = x^2 + 4x - 21
(x-4)/(x-3) - (x+7)/(x+8)
= [(x^2 + 4x - 28) - (x^2 + 4x - 21)]/[(x-3)(x+8)]
= [x^2 + 4x - 28 - x^2 - 4x + 21)]/[(x-3)(x+8)]
= -7/[(x-3)(x+8)]
Now factor the denominator from the second part
x^2 + 5x - 24 = (x-3)(x+8)
The common denominator is again (x-3)(x+8).
Since both terms have the common denominator, you can just add the numerators.
-7/[(x-3)(x+8)] + (x+8)/[(x-3)(x+8)]
= (-7 + x + 8)/[(x-3)(x+8)]
= (x+1)/[(x-3)(x+8)]

Answer 4

it truly is honestly one of the lengthy thanks to attempt this difficulty (you may want to attempt the quadratic equation in case you dare...) yet I went with what popped into my head first. i began by using multiplying that total equation by using 4. That clears out the nasty fraction. then you really're left with 3x^2 + 32x + 20. Now discover 2 numbers that multiply to 240 (3 cases 20) and upload to 32 (the middle time period). I used the field technique because it is large easy and enables you to easily pull out trouble-free factors. i won't be able to fairly draw that yet i'm hoping they're nonetheless training that technique haha. So now you should have 2 linear factors: (3x+20)(x+4). The zeros (or x values) are x = -20/3 and x = -4