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### Re: Schoolwork Help Thread

Posted: **Sun Nov 28, 2010 8:33 pm**

by **noname**

tommyservo7 wrote:noname wrote:
2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.

I tried. Looks like it requires a matrix.

EDIT:

This nifty little thing works excellently. It will teach you how to do it.

Thanks for the help. If this is the way to go, thank you, but I don't this is how it was taught in class. If anyone else has an alternative method, please let me know.

### Re: Schoolwork Help Thread

Posted: **Sun Nov 28, 2010 8:51 pm**

by **AWA**

noname wrote:
2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.

2x+2z=2

5x+3y=4

3y-4z=4

It's a system of equations.

The first equation (solved for z):

z=1-x

Substitute this statement into the third equation:

3y-4(1-x)=4

3y-4+4x=4

3y+4x=8

Now you have a simple system of two variables. Solve with the second equation (equation addition and subtraction):

3y+4x=8

3y+5x=4

-x=4

**x=-4**
Now substitute this x-value into the first equation:

2(-4)+2z=2

-8+2z=2

2z=10

**z=5**
You may choose to use either the x-value or the z-value to solve for y. Let us use the z-value for variety:

3y-4(5)=4

3y-20=4

3y=24

**y=8**
Putting our three results into an ordered triple:

(x,y,z)

**(-4,8,5)**
Check your answer. It should be correct.

### Re: Schoolwork Help Thread

Posted: **Sun Nov 28, 2010 9:06 pm**

by **noname**

AWA wrote:noname wrote:
2x+2z=2

5x+3y=4

3y-4z=4

The textbook says the solution should be (-4,8,5). How do you get this answer? I know it's something really simple, but I can't remember how to do it.

2x+2z=2

5x+3y=4

3y-4z=4

It's a system of equations.

The first equation (solved for z):

z=1-x

Substitute this statement into the third equation:

3y-4(1-x)=4

3y-4+4x=4

3y+4x=8

Now you have a simple system of two variables. Solve with the second equation (equation addition and subtraction):

3y+4x=8

3y+5x=4

-x=4

**x=-4**
Now substitute this x-value into the first equation:

2(-4)+2z=2

-8+2z=2

2z=10

**z=5**
You may choose to use either the x-value or the z-value to solve for y. Let us use the z-value for variety:

3y-4(5)=4

3y-20=4

3y=24

**y=8**
Putting our three results into an ordered triple:

(x,y,z)

**(-4,8,5)**
Check your answer. It should be correct.

That's more like what we learned in class. Thanks to both of you for your help.

### Re: Schoolwork Help Thread

Posted: **Fri Feb 04, 2011 3:57 am**

by **waffley**

My homework will be due before anyone sees this, but here's a couple questions.

This one's multiple choice:

The average velocity of an object for a given motion points to the right. When is it possible for the instantaneous velocity of the object to point to the left? (Select all that apply.)

-at the beginning of the motion

-at some point in the middle of the motion

-at the end of the motion

I just 100% don't get this question and so I got it wrong.

Also this:

You drive in a straight line at 18.0 m/s for 12.0 miles, then at 30.0 m/s for another 12.0 miles.

(a) How does your average speed compare with 24.0 m/s?

speedav < 24.0 m/s

speedav > 24.0 m/s

speedav = 24.0 m/s

Explain.

I originally was like "oh I'm traveling different speeds over the same distance so I can just add those and take the average of that" which ended up being wrong. Later on in the question you calculate the average velocity and I realized that since you are traveling the same distance with different speeds, the time in which you travel those distances is different so you can't really take a shortcut there and you have to actually calculate out the distance traveled over time. Right? Does that make sense?

### Re: Schoolwork Help Thread

Posted: **Fri Feb 04, 2011 11:30 am**

by **ottomagne**

waffley wrote:My homework will be due before anyone sees this, but here's a couple questions.

This one's multiple choice:

The average velocity of an object for a given motion points to the right. When is it possible for the instantaneous velocity of the object to point to the left? (Select all that apply.)

-at the beginning of the motion

-at some point in the middle of the motion

-at the end of the motion

I just 100% don't get this question and so I got it wrong.

Also this:

You drive in a straight line at 18.0 m/s for 12.0 miles, then at 30.0 m/s for another 12.0 miles.

(a) How does your average speed compare with 24.0 m/s?

speedav < 24.0 m/s

speedav > 24.0 m/s

speedav = 24.0 m/s

Explain.

I originally was like "oh I'm traveling different speeds over the same distance so I can just add those and take the average of that" which ended up being wrong. Later on in the question you calculate the average velocity and I realized that since you are traveling the same distance with different speeds, the time in which you travel those distances is different so you can't really take a shortcut there and you have to actually calculate out the distance traveled over time. Right? Does that make sense?

I didn't quite understand the first one, but I do get the second one.

18 m/s = 12/t1, 12/18 s = t1

30 m/s = 12/t2, 12/30 s = t2

Average velocity = dd/dt, or 24/t1+t2

Because t1+t2 > 1 (12/18 is 2/3, 12/30 > 10/30 = 1/3), we know that the average velocity that you traveled will be < 24 m/s.

### Re: Schoolwork Help Thread

Posted: **Sun Mar 20, 2011 10:18 pm**

by **waffley**

Posting this in here because everyone should know about this awesome guy who should probably be a tenured professor at some awesome university.

http://www.youtube.com/user/freelanceteach" onclick="window.open(this.href);return false;

### Re: Schoolwork Help Thread

Posted: **Fri Sep 30, 2011 10:44 pm**

by **petrolika**

Anyone here know anything about Agriculture, and knows how to do DSE (Dry Sheep Equivalent)??