Question

can someone help guide me through solving this?

Answer 1

\[\frac{ 2x+5 }{ x^2-3x-10 }+\frac{ x+1 }{ x+2 }\]

Answer 2

@satellite73 ?

Answer 3

@ganeshie8

Answer 4

\[\frac{ 2x+5 }{ x^2-3x-10 }+\frac{ x+1 }{ x+2 }\] find the old "common denominator" which is probably \(x^2-3x-10\)

Answer 5

that is because \[\frac{ 2x+5 }{ x^2-3x-10 }+\frac{ x+1 }{ x+2 }=\frac{ 2x+5 }{ (x+2)(x-5)}+\frac{ x+1 }{ x+2 }\]

Answer 6

so you have to multiply the second fraction top and bottom by \(x-5\) i.e compute \[\frac{ 2x+5 }{ (x+2)(x-5)}+\frac{ (x+1)(x-5) }{( x+2)(x-5) }\] so make the denominators the same

Answer 7

so you would just mulyiply the second fraction by (x-5) not the first? its just to make the denominators the same right?

Answer 8

then do i cancel out the x-5 on the second fraction?
@satellite73

Answer 9

no no no don't cancel anything!!

Answer 10

the answer to your first question

so you would just mulyiply the second fraction by (x-5) not the first? its just to make the denominators the same right?

is YES

Answer 11

oh ok

Answer 12

then add up in the numerator, leave the denominator the same

Answer 13

don't think about cancelling until you have done all the work
usually nothing cancels

Answer 14

i have \[\frac{ 2+5 }{ (x+2)(x-5) }+\frac{ (x+1)(x-5) }{(x+2)(x-5) }\] so far @satellite73

Answer 15

cept for the missing x

Answer 16

oops sorry yeah that was supposed to be 2x+5

Answer 17

\[\frac{ 2x+5 }{ (x+2)(x-5)}+\frac{ (x+1)(x-5) }{( x+2)(x-5) }\]\[=\frac{2x+5+(x+1)(x-5)}{(x+2)(x-5)}\]

Answer 18

nothing cancels
multiply out in the top, combine like terms

Answer 19

(x^2-2x)/(x+2)(x-5) ?

Answer 20

@satellite73