Question

Hi I am super new to this, I thought I would give it a shot. Can anyone help me solve this: Simplify quantity x squared plus 2 x over quantity 12 x plus 54 end quantity minus quantity 3 minus x over quantity 8 x plus 36

Answer 1

is it like this?
\[\large \frac{x^2+2x}{12x+54}-\frac{3-x}{8x+36} \]

Answer 2

in the first fraction u can factor x in the numerator and 12 in the denominator. and in the second fraction u can factor 4 in the denominator.
can u do it?

Answer 3

sorry. i made a mistake. u cannot factor 12 from the denominator of the first fraction

Answer 4

let's see
\[\large \frac{x^2+2x}{12x+54}-\frac{3-x}{8x+36}=
\frac{\color{red}{x}(x+2)}{\color{red}{6}(2x+9)}-
\frac{3-x}{\color{red}{4}(2x+9)} \]
the LCM is, as u said, \(12(2x+9)\)

Answer 5

so to equal both denominators we multiply the first fraction by 2/2 and the second by 3/3

Answer 6

\[\large \frac{2}{2}\cdot\frac{x(x+2)}{6(2x+9)}-
\frac{3}{3}\cdot\frac{3-x}{4(2x+9)}=
\frac{2x(x+2)}{12(2x+9)}-\frac{9-3x}{12(2x+9)}= \]
\[\large =\frac{2x(x+2)\color{red}{-}(9-3x)}{12(2x+9)} \]

Answer 7

i cannot figure out whay your teacher told u that. it makes no sense at all.

maybe she got a bit confused.

Answer 8

so to finish
\[\large =\frac{2x^2+4x-9+3x}{12(2x+9)}=\frac{2x^2+7x-9}{12(2x+9)} \]
factoring the numerator we got
\[\large =\frac{(2x+9)(x-1)}{12(2x+9)}=\frac{x-1}{12} \]

Answer 9

u r welcome.
i log in this site whenever i have some free time.
if i am not loged in, there is a lot of very intelligent people here. i am sure u will always find someone to help u.

Answer 10

u too have a nice day. bye!