Question

Help me please (x^2 + 2x)/(12x + 54) - (3 - x)/(8x +36)

Answer 1

Hello, your first step is to factor the expression. Can you do that for me, hun?

Answer 2

wait first aren't you supposed to turn the subtraction sign into an addition sign... and you do that by multiplying it by -1 and it only effects the numerator

Answer 3

No - your first step is simplifying.

Answer 4

I actually have to go! Sorry! But I bet @ganeshie8 or @goformit100 can assist you! :)

Answer 5

x(x + 2)/ 6(2x + 9) - (-3 + x)/4(2x + 9)?

Answer 6

you mean

x(x + 2)/ 6(2x + 9) + (-3 + x)/4(2x + 9)?

Answer 7

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

Answer 8

oops ya sorry.. and now I know we're supposed to have a common denominator

Answer 9

yes,
so how to get a common denominator ?

Answer 10

wats lcm of 6 and 4 ?

Answer 11

is it 12 ?

Answer 12

You multiply 4 to the numerator and the denominator of the first fraction and multiply 6 to the numerator and the denominator of the 2nd fraction?

Answer 13

perfect !

Answer 14

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

\(\large \frac{4x(x + 2)}{ 24(2x + 9)} + \frac{6(-3 + x)}{24(2x + 9)} \)

Answer 15

now that we have denominators same,
simply add numerators

Answer 16

ok that makes sense so far

Answer 17

would you multiply the 6 and the 4x in the numerators

Answer 18

yes

Answer 19

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

\(\large \frac{4x(x + 2)}{ 24(2x + 9)} + \frac{6(-3 + x)}{24(2x + 9)} \)

\(\large \frac{4x^2+8x}{ 24(2x + 9)} + \frac{-18 + 6x}{24(2x + 9)} \)

Answer 20

add them now

Answer 21

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

\(\large \frac{4x(x + 2)}{ 24(2x + 9)} + \frac{6(-3 + x)}{24(2x + 9)} \)

\(\large \frac{4x^2+8x}{ 24(2x + 9)} + \frac{-18 + 6x}{24(2x + 9)} \)

\(\large \frac{4x^2+8x-18+6x}{ 24(2x + 9)} \)

Answer 22

simplify

Answer 23

why wouldn't you just multiply the 6 and 4x and just leave it like 24x(x + 2)(x - 3)

Answer 24

could you do that?

Answer 25

dint get u

Answer 26

for the 4x and 6 you distributed but do you have to do that or can you just combine them like 24x(x +2)(-3 + x)

Answer 27

?

Answer 28

you will have to distribute that

Answer 29

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

\(\large \frac{4x(x + 2)}{ 24(2x + 9)} + \frac{6(-3 + x)}{24(2x + 9)} \)

\(\large \frac{4x^2+8x}{ 24(2x + 9)} + \frac{-18 + 6x}{24(2x + 9)} \)

\(\large \frac{4x^2+8x-18+6x}{ 24(2x + 9)} \)

\(\large \frac{4x^2+14x-18}{ 24(2x + 9)} \)

\(\large \frac{2(2x^2+7x-9)}{ 24(2x + 9)} \)

\(\large \frac{2x^2+7x-9}{ 12(2x + 9)} \)

Answer 30

we end up like that
see if it makes some sense

Answer 31

okay so we HAVE to distribute

Answer 32

yup !

Answer 33

see if u can factor numerator
and can cancel something

Answer 34

(2x^2 - 2x)(9x - 9)

Answer 35

\(\large \frac{x(x + 2)}{ 6(2x + 9)} + \frac{(-3 + x)}{4(2x + 9)} \)

\(\large \frac{4x(x + 2)}{ 24(2x + 9)} + \frac{6(-3 + x)}{24(2x + 9)} \)

\(\large \frac{4x^2+8x}{ 24(2x + 9)} + \frac{-18 + 6x}{24(2x + 9)} \)

\(\large \frac{4x^2+8x-18+6x}{ 24(2x + 9)} \)

\(\large \frac{4x^2+14x-18}{ 24(2x + 9)} \)

\(\large \frac{2(2x^2+7x-9)}{ 24(2x + 9)} \)

\(\large \frac{2x^2+7x-9}{ 12(2x + 9)} \)

\(\large \frac{(x-1)(2x+9)}{ 12(2x + 9)} \)

\(\large \frac{x-1}{ 12} \)

Answer 36

that makes soo much sense thank you =)

Answer 37

np :)

Answer 38

wait