Question

Explain in complete sentences and demonstrate how to add:

Answer 1

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

Answer 2

the two fractions are over each other @SolomonZelman

Answer 3

when you add fractions with the same denominator, only the numerator changes while the denominator remains the same. Add like terms in the numerator and you will get two times a number minus four, which is going to be over two times the number.

IDK, ain't good at explaining math using sentences.

Answer 4

could you demonstrate?

Answer 5

\[\frac{x-5}{2x}+\frac{x+1}{2x}~~~~~->\]\[\frac{(x-5)+(x+1)}{2x}\]\[\frac{2x+4}{2x}\]

Answer 6

ohhhh i get it so the denominator doesn't change.

Answer 7

then reduce it,\[\frac{2x+4}{2x}~~~->~~~\frac{2(x+2)}{2x}~~~_>~~~\frac{x+2}{x}\]

And yes the denominator stays the same.

Answer 8

ok i see hold on ill brb

Answer 9

K

Answer 10

k back

Answer 11

I"m here for a couple minutes or so....

Answer 12

ok so thats how you do it

Answer 13

Perform the subtraction: [(x-5)/2x] - [x/(x+3)]
Discuss and identify any possible restrictions that exist in the resulting rational expression.

Answer 14

\[\frac{ x-5 }{ 2x }\]

Answer 15

minus

Answer 16

\[\frac{ x }{ x+3 }\]

Answer 17

demonstrate or sentences?

Answer 18

both, like the first question

Answer 19

I am not good at wrting sentences, but I can show you the math....

Answer 20

ok

Answer 21

\[\frac{x-5}{2x}~-~\frac{x}{x+3}\]\[\frac{x \color{red} {\times (x+3)}}{2x\color{red} {\times (x+3)}}~-~\frac{x \color{red} {\times 2x}}{(x+3) \color{red} {\times 2x} }\]

Answer 22

\[\frac{(x^2+3x)}{(2x^2+3x)}~-~\frac{(2x^2)}{(2x^2+3x)}\]

Answer 23

\[\frac{(x^2+3x)-(2x^2)}{(2x^2+3x)}\]

Answer 24

\[\frac{(-x^2+3x)}{(2x^2+3x)}\]

Answer 25

hmm

Answer 26

so you cross multiply?

Answer 27

no, I find the common denominator, doing it just like with any normal fraction, except that I am finding the common denominator between binomials and variables.

Answer 28

ohhh ok

Answer 29

any possible restrictions that are in the resulting expression?

Answer 30

@SolomonZelman

Answer 31

the denominator cannot equal zero, so \[\color{blue} {2x^2+3x≠0}\]\[\color{blue} {2x^2+3x\color{red} {+0}≠0}\]plug it inside the quadratic formula.
\[\color{blue} {\frac{-3±\sqrt{3^2+(4 \times 2 \times 0)}}{2 \times 2}}\]\[\color{blue} {\frac{-3±\sqrt{9}}{4}}\]\[\color{blue} {\frac{-3±3}{4}}\]\[\color{blue} {x≠ 0~~~~~~~~~~or~~~~~~~~x≠ -1\frac{1}{2}}\]

Answer 32

thanks.