Question

Someone please help me!!!

Answer 1

\[5 \over 2 + x \] plus \[x \over x - 4 \]

Answer 2

@Loser66

Answer 3

@Nerdy_3000

Answer 4

\[\frac{ 5 }{ 2 + x } + \frac{ x }{ x-4 }\] here treat the two terms as fractions.

As any fractions you can add or subtract them only if you have the same denominator.

Here since you do not have the same denominator and the two does not have any common factor we simply multiply them to get the LCD, Least Common Denominator.

So your LCD is (2 + x) ( x -4)

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

Now distribute 5 to (x-4) and distribute x to (2+x) and combine like terms on top and simplify the top. Can you do that?

Answer 5

I don't know how to contribute

Answer 6

@rajee_sam

Answer 7

"distribute", meaning expand

5(x-4) = 5x - 20
x(2 + x) = 2x + x²

now you add all of them because it is added

5x -20 + 2x + x² = x² + 7x - 20

So your fraction is \[\frac{ x ^{2} +7x - 20 }{ (2 + x) (x-4) }\]

Answer 8

Now we see if we can factor the numerator still.

Answer 9

Can you draw it? Cause I don't unertand this

Answer 10

draw what?

Answer 11

The fraction

Answer 12

let me do step by step all at one place so you can see it, if you want explanation for a step see my earlier posts here.

\[\frac{ 5 }{ (2+x )} + \frac{ x }{ (x-4) } \]\[= \frac{ 5(x-4) + x (2 +x) }{ (2 +x) (x-4) }\]

\[= \frac{ (5x - 20) + (2x + x ^{2}) }{ (2 +x) (x-4) }\] \[= \frac{ 5x - 20 + 2x + x ^{2} }{ (2+x) (x-4) }\]\[= \frac{ x ^{2} + 7x - 20 }{ (2 +x) (x-4) }\]

Answer 13

you cannot factor the numerator in a way that it would cancel out with the denominator. So leave it as it is

Answer 14

I meant like this...