(Non-linear) equations involving three or more variables can only be solved in special cases. We shall here consider some of the most useful methods of solution.

**Example 1.**

Solve the following system of equations:

Equation I

Equation II

Equation III

**Solution I:**

From II and III, we get

Plug in ; then this equation becomes .

Also, from I, we get

hence, we obtain , .Thus,we have

, ; and , . Hence, the solutions are

; ; ;

or,

; ; .

**Example 2:**

Solve the following system of equations:

Equation I

Equation II

Equation III

**Solution: 2:**

Write u, v, w for , , respectively; thus

, , …..call this equations A

Multiplying these equations together, we have

. Hence, .

Combining this result with each of the equations in A, we have

; or, ; therefore, we get two sets of linear equations, as follows:

for which the solution set is: ; and the other set is:

for which the solution set if .

**Example 3:**

Solve the following system of equations:

…call this equation (1)

…call this equation (2)

…call this equation (3)

Subtracting (2) from (1), we get the following:

, that is, …call this equation (4).

Similarly, from (1) and (3), we get the following:

…call this equation (5).

Hence, from equations (4) and (5), by division, we get the following:

, hence,

Substituting in Equation (3), we obtain .

From (2), we get the following:

Solving these homogeneous equations (*hint: *put , where m is parameter to be found), we get the following:

, ; and therefore,

or, ; ; and therefore, .

**Example 4:**

Solve the following system of equations:

**Solution 4:**

Multiply the equations by y, z, and x respectively, and add; then,

…call this equation I

Multiply the equations by z, x and y respectively; and add; then

…call this equation II

From (I) and (II), by cross multiplication,

, suppose.

Substitute in any one of the given equations; then,

; hence, we get the following solution:

.

More stuff in the pipeline,

Nalin Pithwa