#1

Solve for Y

x + 5y=15

help?

x + 5y=15

help?

#2

y = (15 - x) / 5

#3

Are you Petarded?

#4

idk if im right... y=3x?

i phail at math

or... y=3-x?

i'll just stop now.

i phail at math

or... y=3-x?

i'll just stop now.

#5

Isolate the 5y by subtracting the x, moving it to the other side and changing its sign. You now have 5y = 15 - x. Divide both sides by 5, and that's yo' answer.

*Last edited by K-Lizzle at Oct 18, 2007,*

#6

That question is too easy to come to The Pit for help.

#7

Y = (15 - X) / 5

#8

Isolate the 5y by subtracting the x, moving it to the other side and changing its sign. You now have 5y = 15 - x. Divide both sides by y, and that's yo' answer.

No, divide both sides by 5 and you solve for y. As says the second post.

#9

Y=2

X=5

X=5

#10

Y=2

X=5

There are infinitely many solutions. You just found one.

#11

HS math ftl

#12

it says solve for Y... so would Y be 2?Y=2

X=5

#13

No, divide both sides by 5 and you solve for y. As says the second post.

That's what I meant. I said y on accident.

Mah bad.

#14

*sigh*

#15

y = -1/5x + 3 <--- As far simplified as I can make it.

unless you have another equation to make a system, then there is not enough information to come up with a number value for y.

unless you have another equation to make a system, then there is not enough information to come up with a number value for y.

*Last edited by Leperaffinity01 at Oct 18, 2007,*

#16

it says solve for Y... so would Y be 2?

No man.

Y = (15 - X) / 5

-_0

You first pass X to the other side, so it changes his sign from + to -. Then you divide everything by 5. So 5Y/5 = Y, and unless you knew the value of X you can't know the final result so you just write the operation [ (15 - X) / 5]

*Last edited by urik at Oct 18, 2007,*

#17

It's pretty simple in my opinion, unless there's a rule I'm missing, any combination of a number + another number multiplied by 5 should give you the same answer...

I'm just using basic algebra substitution.

I'm just using basic algebra substitution.

#18

thank you guys immensely!No man.

Y = (15 - X) / 5

-_0

You first pass X to the other side, so it changes his sign from + to -. Then you divide everything by 5. So 5Y/5 = Y, and unless you knew the value of X you can't know the final result so you just write the operation [ (15 - X) / 5]

#19

I think I learned this in fourth or fifth grade. What grade are you in?

#20

No man.

Y = (15 - X) / 5

-_0

You first pass X to the other side, so it changes his sign from + to -. Then you divide everything by 5. So 5Y/5 = Y, and unless you knew the value of X you can't know the final result so you just write the operation [ (15 - X) / 5]

why do you have the 15-x in parentheses? The 15-x isn't itself a term. The 15 and the x are separate. Keeping them grouped isn't simplified. Unless I'm mistaken

#21

i doubt you learned multi-step equations in fourth grade... i'm in 11th.I think I learned this in fourth grade. What grade are you in?

#22

I haven't been going to school for over 6 months... I must brush up on my math skills, the only thing I learned at school worth keeping.

#23

why do you have the 15-x in parentheses? The 15-x isn't itself a term. The 15 and the x are separate. Keeping them grouped isn't simplified. Unless I'm mistaken

the parentheses are indicating that the entire expression 15-x is divided by 5. If they were not indicated, it would say 15-(x-5) which is not the answer

#24

y = -1/5x + 3 <--- As far simplified as I can make it.

unless you have another equation to make a system, then there is not enough information to come up with a number value for y.

Isn't this the correct answer?

#25

Just finished matrices and linear programming here. Moving on to working with symmetry.At least in pre-calc that is. We're using unit circles to find trigenometric functions in trig

#26

Nah, I remember learning simple algebra at the end of fourth grade and the beginning of fifth grade. That was when I was in India though. Crazy-ass academic system. They were teaching fractions in fifth grade, when I moved to the U.S.

I'm in Linear Algebra now. It's slightly reminiscent of regular algebra, but not really.

I'm in Linear Algebra now. It's slightly reminiscent of regular algebra, but not really.

#27

thank you guys immensely!

Do the operation replacing the letters with numbers to check if we're right.

x + 5y=15

Lets see.

X = 5

(5 + 5x2 = 15)

Y = 2

2 = (15 - 5) / 5

2=2

We're right

#28

y = -1/5x + 3 <--- As far simplified as I can make it.

unless you have another equation to make a system, then there is not enough information to come up with a number value for y.

Fail

X + 5Y = 15

Subtract the X from both side to isolate the Y variable. This is done in all equations that require you to solve for a certain variable.

5Y = 15 - X

Divide by the constant in front of the Y to reduce the variable all the way.

Y = (15 - X) / 5

Separate into two fractions with a denominator of 5. You can do this to any fraction where the variables in the numerator are being added or subtracted, and the denominator is constant.

Y = (15 / 5) - (X / 5)

Simplify.

Y = 3 - X / 5

Done.

NVM, you edited it.

#29

the parentheses are indicating that the entire expression 15-x is divided by 5. If they were not indicated, it would say 15-(x-5) which is not the answer

I know, you'd use the distributive property and essentially multiply the phrase my 1/5. I meant that it wasn't completely simplified. I was always taught to simplify my equations.

#30

Fail

X + 5Y = 15

Subtract the X from both side to isolate the Y variable. This is done in all equations that require you to solve for a certain variable.

5Y = 15 - X

Divide by the constant in front of the Y to reduce the variable all the way.

Y = (15 - X) / 5

Separate into two fractions with a denominator of 5. You can do this to any fraction where the variables in the numerator are being added or subtracted, and the denominator is constant.

Y = (15 / 5) - (X / 5)

Simplify.

Y = 3 - X / 5

Done.

You just used the commutative property, dummy.

They're both the same equation.

#31

I posted right after he edited it. Used to be " - 3 " not " + 3 "

#32

Fail

X + 5Y = 15

Subtract the X from both side to isolate the Y variable. This is done in all equations that require you to solve for a certain variable.

5Y = 15 - X

Divide by the constant in front of the Y to reduce the variable all the way.

Y = (15 - X) / 5

Separate into two fractions with a denominator of 5. You can do this to any fraction where the variables in the numerator are being added or subtracted, and the denominator is constant.

Y = (15 / 5) - (X / 5)

Simplify.

Y = 3 - X / 5

Done.

That's the same thing numbnuts. -1/5x = -x/5 I just put in the one for simplicity & used the communtative property

phail @ phail detection lol

#33

Elementary algebra seems like a good ol' fun time compared to Linear algebra. F*cking number crunching...

#34

Y=2

X=5

that's what i thought!

#35

Elementary algebra seems like a good ol' fun time compared to Linear algebra. F*cking number crunching...

lol

augmented matrices ftl so damn time consuming

#36

Haha, I haven't gotten there yet. I'm at Inter-something Differentiation. First time taking calc while at college. I'll get to Linear Algebra though, seeing as I'm and Aero major.

#37

that's what i thought!

#38

OK, now what you gotta do is go down the road past the old Johnson place. You're gonna find two roads, one parallel and one perpendicular. Now keep going until you come to a highway that bisects it at a 45-degree angle.

Solve for x.

Solve for x.

#39

OK, now what you gotta do is go down the road past the old Johnson place. You're gonna find two roads, one parallel and one perpendicular. Now keep going until you come to a highway that bisects it at a 45-degree angle.

Solve for x.

YEEEEEEAAAAAAAAAAAAAAAAAAAAAAAAH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

sorry couldn't help after seeing your avatar

#40

You don't spend a thousand dollars on clothes... ... that you're never gonna wear...

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DUH! DUH! DUH!

YEEEEEEAAAAAAAAHHHHHHHHH!!!!!!!

DUH! DUH! DUH!