Non-linear System of Equations

[albie09]

I have the concept of solving simulaneous equations. I am trying to use the elimination method to solve the following and am having trouble getting it to check. Where am I messing up? I am labeling the equations with a capital letter and describing the operation performed on the equations prior to giving the results.

Solve and check the following:

(1/x) - (2/y) - (2/z) = 1 (A)
(3/x) + (4/y) + (6/z) = -9 (B)
(1/x) + (2/y) + (2/z) = -5 (C

I eliminated the 'x' term first:

(A-C): -(4/y) - (4/z) = 6 which I simplified to -(2/y) - (2/z) = 3 (D)

(B-3A): (10/y) + (12/z) = -12 which I simplified to (5/y) + (6/z) = -6 (E)

Eliminating the 'z' term:

(3D+E): -(1/y) = 3. Sloving for y: y = -1/3

Eliminating the 'y' term:

(D): -(2/y) - (2/z) = 3
-2/3 - (2/z) = 3. Multiplying thru by 3 and solving for z: z = -6/7
[my first mistake is probably here - a denominator of 7 is tuff. This is a high school text problem]

Solving for the 'x' term:

(A) (1/x) - (2/y) - (2/z) = 1
(1/x) + 2/3 + 12/7 = 1. Multiplying thru by 21 and solving for x: x = -21/29
[I have to have my second mistake here - '-21/29' can't be correct; it make the 'math' too hard]

The 'checks' don't work, so where did I go wrong?

 

[masters]

I have the concept of solving simulaneous equations. I am trying to use the elimination method to solve the following and am having trouble getting it to check. Where am I messing up? I am labeling the equations with a capital letter and describing the operation performed on the equations prior to giving the results.

Solve and check the following:

(1/x) - (2/y) - (2/z) = 1 (A)
(3/x) + (4/y) + (6/z) = -9 (B)
(1/x) + (2/y) + (2/z) = -5 (C

I eliminated the 'x' term first:

(A-C): -(4/y) - (4/z) = 6 which I simplified to -(2/y) - (2/z) = 3 (D)

(B-3A): (10/y) + (12/z) = -12 which I simplified to (5/y) + (6/z) = -6 (E)

Eliminating the 'z' term:

(3D+E): -(1/y) = 3. Sloving for y: y = -1/3

Eliminating the 'y' term:

(D): -(2/y) - (2/z) = 3
-2/3 - (2/z) = 3. Multiplying thru by 3 and solving for z: z = -6/7
[my first mistake is probably here - a denominator of 7 is tuff. This is a high school text problem]

Solving for the 'x' term:

(A) (1/x) - (2/y) - (2/z) = 1
(1/x) + 2/3 + 12/7 = 1. Multiplying thru by 21 and solving for x: x = -21/29
[I have to have my second mistake here - '-21/29' can't be correct; it make the 'math' too hard]

The 'checks' don't work, so where did I go wrong?

Hi albie09,

Looks like you found the y term ok. Rather than shuffle through the rest of what you did, let me show you how I started out. Adding equations (A) and (C) will eliminate the y and z terms, thus allowing you to find x very easily.

(A) \(\displaystyle \frac{1}{x}-\frac{2}{y}-\frac{2}{z}=1\)

(C) \(\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{2}{z}=-5\)
------------------------------------------------------------
\(\displaystyle \frac{2}{x}=-4\)

\(\displaystyle \boxed{x=-\frac{1}{2}}\)

Now, it should be a piece of cake to find z.