You can always do it with a matrix, then work down to row-echelon form (I think that's the correct term). I know most people hate matrices, but that's what calculators are for.

 

in my opinion the easiest method would be to use cramer's rule, as we need operate on 3rd order determinants in this case... try these links ::

www.mathworld.wolfram.com/CramersRule.html or

www.purplemath.com/modules/cramers.htm

 

Speaking of learning, for just two equations elimination and substitution are easier. Find one and replace it in the other equation.

For 3 equations, use Gauss elimination or Cramer's rule, for more than 3 variables, Gauss elimination still applies however cramer's rule might be difficult. Or you could just start grouping equations and solving groups at a time.

Once learnt, learnt! as for applying it could be a tremendous time waster if you plan to do manually in the exams.

 

Dunno if this is already mentions:
where it's
Ax+By=C
Dx+Ey=F .........................for 3 equations
for 2 equations, variables
(F-C*(D/A))/(E-(D/A)*B)->Y
(C-BY)/A->X

Ax+By+Cy=D.........for the rest
for 3 equations,vars:
((DFK+BGL+CHJ-LFC-JGD-KHB)/M)->X
((AHK+DGI+CEL-IHC-LGA-KED)/M)->Y
((AFL+BHI+DEJ-IFD-JHA-LEB)/M)->Z

And plz help me on this:
I got these formula from looking at a program once, but I still have no clue how these formulas are derived from this :
Ax+By=C
Dx+Ey=F

 

Solving simutaneous equations with three variables can be done with the row-echelon matrix way, or the Cramer's Rule if the matrix satisfy certain conditions.

 

Ax+By=C ------1
Dx+Ey=F -- ---2

Multiply 1 by D

ADx+BDy=CD -----3

Multiply 2 by A

ADx+AEy=AF-----4

Subtracting 3 from 4 gives

0x+AEy-BDy = AF - CD

So (AE-BD)y = AF - CD

y = (AF - CD)/(AE - BD)

To solve for x, substitute this value for y in your fourth equation.

Now to express y in the form you wrote above, divide the top and bottom lines by A

y = {F - C(D/A)}/{E - (D/A)B}

For 3 variables, you do the same but you need more steps.

Firstly, reduce the three 3 variable equations to two 2 variable ones by eliminating one variable. Then do what I did above to solve for the 2 remaining variables.

Then you can substitute these into one of the original equations and solve for the third variable.

__________________
Len

 

I would say that if you know there exist unique solutions then probably the easiest way solve the set of equations is Cramer's rule

 

i will go for cramer rule to solve these problems

 

TI-83 is the way to go! Engr undergrads surely know how to add and subtract by now. In your case, it's a big waste of time to work it out algebraically.

 

do not worry, he will use it after graduation, but for know he needs to know how the things can be done, and how to apply the rules.

__________________
Its what your friend in Your mind, what you in your friends mind

 

He has probably already graduated as he asked the question 3 years ago.

Mike.

 

This is a standar problem with first year students in university here in Australia even though it is supposed to be taught before hand. I have to often teach this as an extra, yet when they see it students tell me "it is simple" then discover how to manipulate formula.

Look outside the square then square it