Question

Simplify this equation and state any restrictions on the variables.
(x+1)/((x^2)+2x-8)-(x)/(4x-8)

Answer 1

\[\frac{ x+1 }{ x ^{2}+2x-8 }- \frac{ x }{4x-8 }\]

Answer 2

@jim_thompson5910 can you help me with this?

Answer 3

x^2 + 2x - 8 factors to ???

Answer 4

i don't remember how to factor

Answer 5

@jim_thompson5910

Answer 6

@Maaf1234
@Dodo1
@zepdrix
@mathstudent55

Answer 7

Find two numbers that multiply to -8 and add to 2 at the same time

these two numbers are -2 and 4

Answer 8

so it factors to (x-2)(x+4)

Answer 9

4x-8 factors to 4(x-2)

Answer 10

this means

\[\large \frac{ x+1 }{ x^2+2x-8 }- \frac{ x }{4x-8 }\]

turns into

\[\large \frac{ x+1 }{ (x-2)(x+4) }- \frac{ x }{ 4(x-2) }\]

Answer 11

now you need to get the denominators the same

Answer 12

the LCD is 4(x-2)(x+4)

\[\large \frac{ x+1 }{ (x-2)(x+4) }- \frac{ x }{ 4(x-2) }\]

\[\large \frac{ 4(x+1) }{ 4(x-2)(x+4) }- \frac{ x }{ 4(x-2) }\]

\[\large \frac{ 4(x+1) }{ 4(x-2)(x+4) }- \frac{ x(x+4) }{ 4(x-2)(x+4) }\]

\[\large \frac{ 4(x+1) - x(x+4) }{ 4(x-2)(x+4) }\]

I'll let you finish

Answer 13

@jim_thompson5910
so would it be
\[\frac{ 6x+4 }{ 4x+4 }\]

Answer 14

no

Answer 15

\[\large \frac{ 4(x+1) - x(x+4) }{ 4(x-2)(x+4) }\]

\[\large \frac{ 4x+4 - x^2-4x }{ 4(x-2)(x+4) }\]

\[\large \frac{ 4 - x^2 }{ 4(x-2)(x+4) }\]

keep going

Answer 16

\[\frac{ 4-x^{2} }{ 4x-4}\]

Answer 17

@jim_thompson5910

Answer 18

\[\large \frac{ 4 - x^2 }{ 4(x-2)(x+4) }\]

\[\large \frac{ -(x-2)(x+2) }{ 4(x-2)(x+4) }\]

\[\large \frac{ -(x+2) }{ 4(x+4) }\]

\[\large -\frac{ x+2 }{ 4x+16 }\]

Answer 19

Omg so my answer is wrong smh

Answer 20

well just keep practicing and you'll get better

Answer 21

Is this it or it there more?

Answer 22

nope that's it

Answer 23

thank you

Answer 24

yw