Question

Help with adding rational expressions?

Answer 1

So the first thing I did was factor b^2 + 5b + 4 which was (b+4)(b+ 1)
That was my least common denominator so I had to get b/b+1 and 7/b+4 to that so I multiplied b/b+1 by b + 4 to get b(b + 4)/(b + 4)(b + 1) and then I multiplied 7/b+4 by b + 1 to get 7(b + 1)/(b + 4)(b + 1)..not sure where to go from there?

What I have

-3 b(b + 4) 7(b + 1)
_____________ + _____________ = ______________
(b + 4)(b + 1) (b + 4)(b + 1) (b + 4)(b + 1)

Now what?

Answer 2

now notice that the common denominator gets cancelled,
like a/b=c/b gives a=c
in same way, a/b+c/b=d/b gives a+c = d
got this ?

Answer 3

Not really..that's confusing for me

Answer 4

ok,
when denominators are common, we combine the numerator ,
\(\Large \dfrac{☺}{♥}+\dfrac{♦}{♥}=\dfrac{☺+♦}{♥}\)
right ?
so what about your left side ? what will it become ?

Answer 5

So it will become b^2 + b/(b+4(b+1)
sorry about the wait, OS wouldn't load so I had to switch to IE

Answer 6

it would not load for me too!
and then numerator of left side is just -3+b(b+4)
since denominator's are common, we can equate the numerators,
so can u solve
-3+b(b+4) = 7(b+1)
for b?q

Answer 7

ohh now we solve I get it, one sec

Answer 8

you'll have to solve a quadratic in b :)

Answer 9

could u get 2 values of b ?

Answer 10

Hartnn, you have the funniest explanations.

Answer 11

I've been trying to do it but it's not really working for me :/

Answer 12

I got b^2 - 7b = 6

Answer 13

-3+b(b+4) = 7(b+1)
-3+b^2+4b = 7b+7

b^2+4b-7b = 7+3

got this ?

Answer 14

Ohh, I didn't multiply 4 by b..okay I got that..what ahappened to the 7b on the left? and how come it's 7 + 3??

Answer 15

-3+b^2+4b = 7b+7
Add 3 on both sides, and subtract 7b from both sides.
b^2+4b-7b = 7+3

Answer 16

see whether you get the quadratic equation as
b^2-3b-10-0

Answer 17

ohhh okay

Answer 18

i meant
b^2-3b-10=0

can you solve it ? and did you get this on your own too ?

Answer 19

Hmm
b^2-3b-10=0
+ 10 on both sides?

b^2-3b = 10
add 3b to both sides?
b^2 = 3b + 10
sqrt both sides?
b = sqrt 3b + 10?

Answer 20

why !?
ok, can you tell me 2 numbers whose sum is -3 and product is -10 ?

Answer 21

-5 and 2, or -2 and 5?

Answer 22

-2 and 5 will not give u -3

so, -5 and 2 it is!
so we split -3x = -5x+2x

x^2-5x+2x-10=0

can you factor out x from 1st 2 terms and 2 from last 2 ? what would u get ?

Answer 23

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

Answer 24

correct!
now factor out x-5 from both terms :)

Answer 25

so...x + 2?

Answer 26

or by factor out you mean make them back to what we had before?

Answer 27

here are the steps, have a look
\(x(x-5)+2(x-5)=0 \\ (x-5)(x+2)=0 \\ x-5 =0 \: or \:x+2=0\)
so what are the 2 values of x we get ?

Answer 28

x = 5 or x = -2?

Answer 29

yes, thats your final answer! :)
ask if any doubt in any step :)

Answer 30

Can I show you the steps and then you tell me if I wrote them down correctly? I have more of these and I want to see if I can do them with the method we used before I come back

Answer 31

ofcourse!
i will verify each step, and will tell if you missed any step :)

Answer 32

Okay! thanks! So first
we had
-3 b 7
______________ + ______ = ______
b^2 + 5b + 4 b + 1 b + 4
then we factored the first one into (the least common demoninator)

-3
______________
(b + 4)(b + 1)
and we need the rest to have l multipled the second one by (b + 4) and the third one by (b + 1) so now we had

-3 b(b + 4) 7(b + 1)
______________ + ______________ = ____________
(b + 4)(b + 1) (b + 4)(b + 1) (b + 4)(b + 1)

and after thatwe had
-3 + b(b + 4) = 7(b + 1)
-3 + b^2 + 4b = 7b + 7
+3 on both sides
b^2 + 4b = 7b + 10
-7b on both sides
b^2 - 3b = 10
-10on both sides
b^2 - 3b - 10 = 0
factored the left side
b(b - 5) + 2(b - 5) <--confused about this and on since we had x instead and we had (x + 2)?

Answer 33

-3 b(b + 4) 7(b + 1)
______________ + ______________ = ____________
(b + 4)(b + 1) (b + 4)(b + 1) (b + 4)(b + 1)

and after thatwe had
***ADD a step here,
-3+ b(b + 4) 7(b + 1)
___________ = ____________
(b + 4)(b + 1) (b + 4)(b + 1)
****

-3 + b(b + 4) = 7(b + 1)


...
b^2 - 3b - 10 = 0
b^2-5b +2b -10=0

basically just replace x by 'b'

Answer 34

b^2 - 3b - 10 = 0
b^2-5b +2b -10=0
b(b-5)+2(b-5)=0
(b-5)(b+2)=0

b-5=0 or b+2=0
b=5 or b=-2
b=5,-2

Answer 35

I don't understand where the (b + 2) came in though, did you just take the b that was outside of (b - 5) and the +2 outside of the other (b - 5)?

Answer 36

exactly!
its like
\(\Large \alpha (stuff)+\beta(stuff)= (\alpha +\beta)(stuff) \)

Answer 37

here, stuff = x-5

Answer 38

thats called as factoring

Answer 39

ooh okay, I'm going to write it down in my book , thanks hartnn!

Answer 40

welcome ^_^

Answer 41

Wait one question, for the answer do I need to include the numbers that b cannot be in the original equations to keep them from being 0 in the demoninator?

Answer 42

yes, sometimes domain is explicitly asked, but even if its not asked, its good to write it, so mention that which values b cannot take, can you find them ?

Answer 43

b couldn't be - and -4?

Answer 44

-1*

Answer 45

correct! :)
and your reason is also correct, denominator cannot be 0 .
good!

Answer 46

Awesome! Thank you!!