Problem

[patricialec]

Hi,
I'm trying to figure out a problem like this:

3X - 2X + 3Z = 11
2X +3Y + Z = 3
5X + 14Y -Z = 1

:shock: :shock:
and I'd like to know where to start first. I think I should combine like terms (is there a particular way to do this) and I also think I should multiply the problem by a number to combine like terms....can someone get me started on this....I would appreciate any help..

 

[wjm11]

3X - 2X + 3Z = 11
2X +3Y + Z = 3
5X + 14Y -Z = 1

and I'd like to know where to start first. I think I should combine like terms (is there a particular way to do this)

When faced with 3 equations and 3 unknowns, matrices is a great tool. Are you familiar with Gaussian elimination? Here are a few sites worth checking out; they provide examples:

http://www.sosmath.com/matrix/system1/system1.html
http://en.wikipedia.org/wiki/Gaussian_elimination
http://ceee.rice.edu/Books/CS/chapter2/linear43.html

There are many more.

 

[Denis]

and I'd like to know where to start first. I think I should combine like terms (is there a particular way to do this) and I also think I should multiply the problem by a number to combine like terms....can someone get me started on this....I would appreciate any help..

Patricia, you tell us something like this with most of the problems you post.
Are you a student attending classes or learning by yourself?
Important for us to know, as it makes it easier to help...

 

[Deleted member 4993]

Hi,
I'm trying to figure out a problem like this:

3X - 2X + 3Z = 11 ? I think that is incorrect
2X +3Y + Z = 3
5X + 14Y -Z = 1

:shock: :shock:
and I'd like to know where to start first. I think I should combine like terms (is there a particular way to do this) and I also think I should multiply the problem by a number to combine like terms....can someone get me started on this....I would appreciate any help..

 

[mmm4444bot]

Gauss-Jordan Elimination is easy, if you know what you're doing.

I'll suggest a different approach for consideration.

I noticed in your system of three equations (I agree with Subhotosh that your post contains at least one typographical error in the equations) that the coefficients on Z in the second and third equation are equal and opposite.

In other words, I see a +Z in one equation and a -Z in another equation.

Adding these two equations together eliminates Z because Z - Z is zero. The result of adding the second and third equation is a new equation that contains only X and Y.

Next, if you multiply both sides of the third equation by 3, you'll get a new equation where the coefficient on Z is -3.

This new equation can be added to the first equation, and (again) Z will be eliminated from the result because 3Z - 3Z is zero.

Now you'll have two new equations in X and Y only. In other words, you will have reduced the problem to solving a system of two equations in two unknowns, followed by easy substitution to complete the solution set.

Solving for X and Y from two equations is easy. Correct?

Once you find the values of X and Y, you can then substitute them into any of the three original equations, and solve for Z.