Question

algebra

Answer 1

y=5/3x+8

Answer 2

Okay,
\[y+2 = \frac{5}{3}(x+6)\]You want \[?x + ?y = ?\]
Any ideas how to start?

Answer 3

x=3y5−245

Answer 4

y=8+5x/3

Answer 5

Why don't we get rid of that fraction first? Multiply everything by the denominator of the fraction:
\[3*y + 3*2 = 3*\frac{5}{3}(x+6)\]\[3y + 6= 5(x+6)\]
Can you rearrange that so that the \(x\) and \(y\) terms are on the left, and anything else on the right?

Answer 6

y=5/3x+8

Answer 7

Why do you keep typing that stuff?

Answer 8

Do you know what the distributive property is?

Answer 9

im lost

Answer 10

Do you know what the distributive property is?

\[a(b+c) = a*b + a*c\]

it's how you get rid of parentheses like we have in
\[3y+6=5(x+6)\]

Can you use it to get rid of the parentheses, please?

Answer 11

y=13 x

Answer 12

Come on...you had \[3y+6 = 5(x+6)\]and you are telling me that becomes \(y = 13x\)
I want you to do just one simple step: use the distributive property to remove the parentheses. Do not simplify, do not collect like terms, leave everything there exactly as it is if you do only that one step.

Answer 13

3y+6=5x+6

Answer 14

Well, closer, but no.

\[5(x+6) = 5*x + 5*6 = 5x + 30\]

So, \[3y + 6 = 5(x+6)\]\[3y+6 = 5*x + 5*6\]\[3y+6 = 5x +30\]

Now, can you rearrange that so that the terms with \(x\) and \(y\) in them are on the left-hand side, and the terms with only numbers are on the right-hand side?

Answer 15

?????

Answer 16

Please do what I ask. We'll save a lot of time if you do that.

\[3y+6 = 5x+30\]Rearrange that so that you have an equation with the terms containing \(x\) or \(y\) on the left side of the = and any terms not containing one of them on the right side of the =.

Answer 17

Let's look at the first term: \(3y\)

Do we want that on the left-hand side or the right-hand side?

Answer 18

+30 its wrong

Answer 19

please medal as I spent lots of time in this

Answer 20

@MD152727 that is NOT what the problem asks you to do.

Answer 21

problem wants the equation in standard form:
\[Ax + By = C\]

Answer 22

:? *facepalm*

Answer 23

Yeah, I know the feeling :-)

Answer 24

My gosh!

Answer 25

Helpful tip that has saved me countless times over the years: when you think you have an answer, check your work. When you have checked and are sure that you didn't make a stupid mistake somewhere, re-read the problem and make sure you have answered the question asked by the problem, and not some other question.

Answer 26

That is what I understood! I am an 11 years old in 9th grade

Answer 27

I'm still not get the problem solve

Answer 28

ok, sorry ignore me, i annoyed you two

Answer 29

??

Answer 30

Perhaps if you did as I requested, @EKKERKING, we could be done...

I will ask again: the term \(3y\), do we want to keep it on the left side or move it to the right side?

Answer 31

move it

Answer 32

Let me remind you of my instructions:

Rearrange that so that you have an equation with the terms containing x or y on the left side of the = and any terms not containing one of them on the right side of the =.

Do we want to move \(3y\) to the right side, or keep it on the left side?

Answer 33

leave it like that

Answer 34

Excellent! That's correct.

Here's what we have so far, just so we can see it:
\[3y + 6=5x+30\]We decided that the \(3y\) stays on the left.

How about the \(+6\), keep on the left, or move to the right?

Answer 35

move it to the right

Answer 36

Very good. How do we do that?

Answer 37

3y+5x=6+30

Answer 38

No, I asked "how?"

\[3y+6 = 5x+30\]If we to move that \(+6\) out of the left-hand side, we must subtract it from both sides.
\[3y+6 -6 = 5x + 30 -6\]\[3y = 5x + 24\]
Any question about that?

Answer 39

bro I'm only missing 1 number

Answer 40

Do you understand what I did there?

Answer 41

no

Answer 42

What grade are you in?

Answer 43

7

Answer 44

help me with that number please

Answer 45

Just curious.

\[3y + 6 = 5x + 30\]You can do anything you want, pretty much, as long as you do it to both sides of the equation.

We have a \(+6\) on the left side, but we don't want it there. To get rid of it, we subtract \(6\) because \(+6 - 6 = 0\). But to subtract it from the left side, we also need to subtract it from the right side or we won't have a true equation any longer.
\[3y + 6 - 6 = 5x + 30 -6\]\[3y +\cancel{6 - 6} = 5x + 30 - 6\]\[3y = 5x + 30 - 6\]But \(30-6 = 24\) so we can simplify the right hand side as well:
\[3y = 5x +24\]

Answer 46

Now, how about that \(5x\) term: keep it where it is, or move it to the left side?

Answer 47

it was -24 thanks buddy

Answer 48

Good luck!