Question

@satellite73 @hilbertboy96 i could use some help , if you have enough time of course.... (i will post it below)

Answer 1

I got time

Answer 2

Well what you want to do is isolate the variable as much as you can that you need to solve so

on the left since j is on the left, 7k, 10l and 4 have nothing to do with j so move them to the other side and you got 8jkl= whatever you get from the first step. You then divide by everything else but j to get j on its own.

Answer 3

don't you hate math?

Answer 4

so what would the answer be (sry i usually try to get some visuals in my mind) @satellite73 I do not hate math, i am usually extremely good at it, but i just don't "enjoy" math its too stressful for my liking

Answer 5

\[\huge 9\color{red}jkl +7k+10l+4=-9k-3l+8\]

Answer 6

Well as i said when you move that stuff to the other side you get....

\[8jkl = -16k - 13l + 4\]

Divide by 8kl and you get

\[j = \frac{ -16k -13l +4 }{ 8kl }\]

(First choice)

Answer 7

subtract \(7k\) and subtract \(10l\) and subtract \(4\) from the left hand side

Answer 8

and you get what @hilbertboy96 wrote

Answer 9

Yup. I am just a guy who can solve math in his head quick

Answer 10

I kind of get it... (sorry im just so new to this, ive never struggled like this before hah)

Answer 11

here is another one (i have a few) i will try to solve it also and then we can compare answers k? if i get stuck i will tell you

Answer 12

Ok

Answer 13

Just let me know if you get stuck

Answer 14

ughhhh im lost x_X

Answer 15

Haha ok. I wrote this one out for myself so I will show and explain the steps.

We have.....
\[6fg + 6fh - 4f -8 = -9g + 8\]

This is a tougher one to solve for f compared to the last one since there are more then one term with f in it correct?

Answer 16

yes, i got lost lol

Answer 17

That is fine lol. OK we got one term that has no variable so that moves to the right and we get....

\[6fg + 6fh - 4f = -9g + 16\]

Can you think of what we might need to do next?

Answer 18

We got three terms and each one has f in common so what can we do knowing this?

Answer 19

Or is this what is confusing and losing you?

Answer 20

im just lost all together haha
uhmm would we carry the 6g and 6h to the other side ??????

Answer 21

we can not do that since the f would go with it and then we got f on both sides. We can factor f out like so....

\[f(6g + 6h -4) = -9g + 16\]

You see how I did that and why I can?

Answer 22

yes i understand how you did that

Answer 23

ok now we divided everything that is multiplied by f over and we solve f....

\[f = \frac{ -9g + 16 }{ 6g + 6h - 4 }\]

This is one of your choices correct?

Answer 24

yess, wow that actual seemed really simple lol xD , are you busy i have a few more?

Answer 25

I am fairly unbusy and bored. So throw what you got at me

Answer 26

i got textbooks, and pencils, and notepads should i go on and throw those at you ? lol

Answer 27

Sure! lol

Answer 28

Yess!! lol here is the prob i will try to solve also

Answer 29

Shall give you hint. Follow same format as last problem

Answer 30

okay i will

Answer 31

p(-6q+5r-10)= 6q-9 ?? am i on the right track or no?

Answer 32

nope you need to do one step first then factor. Add the term without the p to the other side to get -6pq + 5pr -p = 6q + 1

Then factor p out of the left and divide

Answer 33

so i should do your step first :

-6pq + 5pr -p = 6q + 10

then my step:

p(-6q+5r-p)= 6q- 19 (or should i put "9+10 ??? )

equaling in

???

Answer 34

Ok I will do as i did before by showing steps and explaining

\[-6pq + 5pr - p +10 = 6q - 9\]

We subtract ten to both sides to get

\[-6pq + 5pr -p = 6q + 1 \] (since -9 + 10 = 1 on the right)

Now we factor p out

\[p(-6q + 5r -1) = 6q + 1\] (and notice the -1 on the left is because if you mutply the p back in that is how you get the -p)

Now divide and get

\[p = \frac{ 6q + 1 }{ -6q + 5r + 1 }\]

Answer 35

Ahhh crud xD first step in subtracting 10 and also the factoring i messed up so final answer should be

\[p = \frac{ 6q - 19 }{ -6q + 5r - 1 }\]

Answer 36

Well nott the factoring just lost my minus when transfering my answer

Answer 37

okay yea i was gonna say something, but then again i ddn't want to confuse you since you know this way better than me lol

Answer 38

Even the best people at math make simple mistakes

Answer 39

and you actually messed up again lol 9-10 = -1 so it would be -6q-1 (for the numerator) (and the site im on can prove it lol)

Answer 40

no it is -19 since you got -9 and you are subtracting 10 so you have -9-10 = -19

Answer 41

your wya makes more sense but can i show you the explanation they gave me (it only shows me after I answer lol)

Answer 42

ahhh ok i see what they are doing....ok let me redo my work for you lol

Answer 43

and yes also i transferred the problem wrong xD

Answer 44

here is the mess they show me afterwards lol

Answer 45

\[-6pq + 5pr - p - 10 = 6q - 9\]
\[-6pq + 5pr - p = 6q + 1\]
\[-p(6pq-5pr+1) = 6q + 1\]
(the reason for -p is because they are preventing a negative in the denominator which is a rule that you can not have)
\[-p = \frac{ 6q + 1 }{ 6q - 5r +1 }\]
\[p = \frac{ -1(6q +1) }{ 6q - 5r + 1 }\]
\[p = \frac{ -6q -1 }{ 6q -5r +1 }\]

Answer 46

yea basically what I got lol

Answer 47

but i just put the negative in the top a different way that works

Answer 48

and in third step suppose to be no p's in the parenthesies xD

Answer 49

well that one was weird one more ques (of this topic) thank you so much for helping me!!!

Answer 50

yep anytime!

Answer 51

Ok this one I will help you through since the last ones were confusing

Answer 52

ohhh this is one of those weirdy ones ahhh, i wish to be an astronomer or cosmologist but i will never make it if i can't understand this math :'(

Answer 53

You will make it. Practice is all you need and guidance. Ok first I will write out the original

\[\frac{ -2r + 7s }{ 3 } = \frac{ 4r + 4t }{ 6 }\]

We got two fractions here which is very unconventional!

So first thing we want to do is get rid of these fractions by multiplying both sides by 6 since that will cancel the 6 out on the right and the 3 on the left as you will see....

\[\frac{ 6(-2r + 7s) }{ 3 } = \frac{ 6(4r + 4t) }{ 6 }\]

The 6 over 6 on the right cancels but the 6 and 3 on the left makes the 3 disappear and the 6 becomes a 2 since you are dividing it by 3.....

\[2(-2r + 7s) = 4r + 4t \]

Up to here does this make sense?

Answer 54

yes that makes sense i was on the right track about multiplying by 6 becuase that gets rid of the fractions all together,

Answer 55

correct but you got that 2 left on the left. We can do this two ways but this way is simpler. Leave the 2 not multiplied through and remove a 2 on the right

\[2(-2r + 7s) = 2(2r + 2t)\]

The twos that are pulled out cancel and you get

\[-2r + 7s = 2r + 2t\] does this make sense what I did?

Answer 56

so in a sense, are you saying if we were to "multiply" the left , then we should "multiply" the right and then even though we do not work it out, we can just cancel them out (hope that made some sense lol)

Answer 57

yes since they both got 2 in common when i pulled that one out on the right they can cancel. Now to finish the problem...

\[-4r = 2t - 7s\]

Now we are dividing by a negative again so what I am going to do is divide by just positive 4 and multiply a -1 through the top.

\[r = \frac{ -1(2t - 7s) }{ 4 }\]
\[r = \frac{ -2t + 7s }{ 4 }\](Choice c)

Answer 58

oh i see how you did that ! well i passes that topic fwew i only have 2 topics left should equal out to about 10 probs is there any way you could help me, i promise im not cheating your answers or anything im taking notes and everything !!!

Answer 59

I can keep helping lol and I believe you!

Answer 60

the next one is formulas, im decent at them ,but the more coplex ones i struggle ... fwew Lets do this !!!

Answer 61

ok so the volume of a sphere is

\[V = \frac{ 4 }{ 3 } \pi r^3\]

Ad we want to get r by itself. Let me ask you first. What would you do first?

Answer 62

Uhm I think i would try to somehow divide "V" by the 4/3 , but im not sure thats right??? (this one is tricky for me)

Answer 63

well we can divide V by 4/3 yes but an easier way so you dont have two fraction bars is multiply by 3 first then divide by 4 and the pi since pi is just a number.

\[V = \frac{ 4 }{ 3 } \pi r^3\]
\[3V = 4 \pi r^3\]
\[\frac{ 3V }{ 4\pi } = r^3\]

Now we need to take a cube root but your answer needs to be in a power form for the root. The rule for roots and powers is



So to take a cube root in power form we essentially take the right and left to the power of 1/3

\[(\frac{ 3V }{ 4\pi })^{1/3} = r\]