Question

Please help by explaining this to me? :)

Answer 1

the denominator cannot be zero
so \(x+5\neq 0\) meaning \(x\neq -5\)

Answer 2

Okay, you're referring to these types of expressions? Not just this specific problem, right? How would you solve this problem?

Answer 3

Haper that is how you solve for the restrictions on x.
It looks like the rest of your question got cut off, `and what` is all i see in the picture.

Answer 4

those value(s) represent.***

Answer 5

My teachers tend to want things done step by step and for me to explain as much as possible but I don't know how to "properly" solve this problem. I solved all the others but none of them, I had to explain :\

Answer 6

Soooo I would start by multiplying through by the LCD, which in this case appears to be 6(x+5).

Answer 7

And that will give us thisssss,
\[\Large 6(x+3)+x(x+5)=5(x+5)\]
I skipped a bit of the simplification there, so lemme know if that was confusing.

Answer 8

I'm still confused :\

Answer 9

\[\Large \frac{x+3}{x+5}+\frac{x}{6}=\frac{5}{6}\]Let's try multiplying by each denominator one at a time, maybe that will make more sense to you :o

So let's start by multiplying both sides by 6.\[\Large 6\left(\frac{x+3}{x+5}+\frac{x}{6}\right)=\left(\frac{5}{6}\right)6\]

Answer 10

On the left, we distribute the 6 to each term in the brackets,\[\Large \frac{6(x+3)}{x+5}+\frac{6x}{6}=\frac{5\cdot6}{6}\]

Answer 11

So then we get some nice cancellations,\[\Large \frac{6(x+3)}{x+5}+\frac{\cancel6x}{\cancel6}=\frac{5\cdot\cancel6}{\cancel6}\]

Answer 12

\[\Large \frac{6(x+3)}{x+5}+x=5\]So that cleaned things up a bit for us, right? :o

Answer 13

Very much :D

Answer 14

So we still have another denominator to deal with :O
Repeat this process with x+5.

Answer 15

\[\Large (x+5)\left(\frac{6(x+3)}{x+5}+x\right)=\left(5\right)(x+5)\]

Answer 16

\[\Large \frac{6(x+5)(x+3)}{x+5}+x(x+5)=5(x+5)\]

Answer 17

and then cancel some stuff :o

Answer 18

You cancel out the x + 5 so 6(x + 3) + x(x + 5) = 5(x + 5)... and thats it?

Answer 19

Do you multiply the numbers?

Answer 20

Yah we have to multiply out some stuff from there.

Answer 21

\[\Large 6(x+3)+x(x+5)=5x+25\]What do you get on the left side? :)

Answer 22

6x + 18 + 2x + 5x

Answer 23

So 13x + 18? Please tell this is correct and that I'm not going crazy here

Answer 24

Woops! x*x is not 2x! :O

Answer 25

It's x^2? I keep getting that confused apparently because I thought it was x^2 but someone told me it was 2x -.-

Answer 26

\[\Large 2x=x+x\qquad\qquad\qquad x^2=x\cdot x\]

Yah there is a difference :D heh

Answer 27

Hmm the rest of it looks good though :o
\[\Large 6x + 18 + x^2 + 5x=5x+25\]

Answer 28

Can you please continue, mon? I think I'm getting it but I want to see if I'm correct since I have two theories :)

Answer 29

Subtracting 5x from each side, and rearranging some stuff, gives us,\[\Large x^2+6x+18=25\]

Answer 30

Subtracting 25 from each side,\[\Large x^2+6x-7=0\]

Answer 31

From here we can either use the Quadratic Formula or factor the polynomial if possible.

Answer 32

Looks like it factors in this case.
Do you see the factors? :)

Answer 33

7, -1 :)

Answer 34

\[\Large (x+7)(x-1)=0\]Yah that sounds right.

Answer 35

Understand how to solve for x from here? :o

Answer 36

Yes! :)

Answer 37

you can also solve it by transposing x/6 to R.H.S
\[\frac{ x+3 }{x+5 }=\frac{ 5 }{ 6 }-\frac{ x }{ 6 }=\frac{ 5-x }{6 }\]
cross multiply
\[6x+18=-\left( x-5 \right)\left( x+5 \right)=-\left( x ^{2}-25 \right)\]
\[x ^{2}-25+6x+18=0,or x ^{2}+6x-7=0\]
further you have the solutions as given by friends.