How would you describe the method of elimination for solving a system of three linear equations in three unknowns?
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Let the 3 equations be A, B and C, containing the unknowns x, y and z.
Pick any 2: say A and B.
Re-arranged A so that it reads: z = some expression containing only x and y.
Call this new equation A2.
Repeat for B: z = a different expression containing only x and y.
Call this new equation B2.
So now you can write: z = equation A2 = equation B2.
Thus z is eliminated.
Call this equation N1: it only contains x and y.
Now repeat all of that from the top, but with a different pair: say A and C
You end up with another equation, N2, that also only contains x and y.
N1 and N2 are now a pair of simultaneous equations in two unknowns: x and y.
Solve them for x and y, producing x1 and y1.
Now substitute these values, x1 and y1 in any one of the original 3 equations: A, B or C.
Hence, solve for z, producing z1.
Check your result for z by substituting x1 and y1 in any other one of the other original 3 equations: A, B or C.
You should obtain the same value of z1.
That, in words, is the answer.
An alternative method is to use determinants - but that is difficult to describe here with this very restrictive, non-graphical, text-only format that Yahoo insists on keeping here in a Mathematical Section of all places!
Pick any 2: say A and B.
Re-arranged A so that it reads: z = some expression containing only x and y.
Call this new equation A2.
Repeat for B: z = a different expression containing only x and y.
Call this new equation B2.
So now you can write: z = equation A2 = equation B2.
Thus z is eliminated.
Call this equation N1: it only contains x and y.
Now repeat all of that from the top, but with a different pair: say A and C
You end up with another equation, N2, that also only contains x and y.
N1 and N2 are now a pair of simultaneous equations in two unknowns: x and y.
Solve them for x and y, producing x1 and y1.
Now substitute these values, x1 and y1 in any one of the original 3 equations: A, B or C.
Hence, solve for z, producing z1.
Check your result for z by substituting x1 and y1 in any other one of the other original 3 equations: A, B or C.
You should obtain the same value of z1.
That, in words, is the answer.
An alternative method is to use determinants - but that is difficult to describe here with this very restrictive, non-graphical, text-only format that Yahoo insists on keeping here in a Mathematical Section of all places!
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You are using a combination of substitution and adding.