Introduction to solving
systems of non linear equations:
Non linear system is a system which is not linear, that is, a
system which does not satisfy the superposition principle, or whose output is
not directly proportional to its input. Less technically, a nonlinear
system is any problem where the variable(s) to be solved for cannot be written
as a linear combination of independent components. A no homogeneous system,
which is linear apart from the presence of a function of the independent variables, is nonlinear according to a strict definition, but such systems are
usually studied alongside linear systems. Let us discuss about solving systems
of non linear equations.
Non Linear Equations on Solving Systems of Non Linear Equations:
Solving System of Non
linear Equations are as follows:
• Non linear equations do not produce a straight line and the exponent is to
the second power or more.
• X2+4Y =6 is an example of a nonlinear equation.
• Equation whose graphs do not shape a straight line (linear) is called a
Nonlinear Equation.
• In a nonlinear equation, the variables are also quantity better than 1 or
less than 1, but never 1.
Examples of System of
Non linear Equations:
• 3x2 + 4y - 2 = 0 is a nonlinear equation.
• x3 + 3x2 - 5xy - 2 = 0 are the exemplar of nonlinear
equation.
• X1/3+Y1/3 is a nonlinear equation.
Example on Solving Systems of Non Linear Equations:
Solving Example for
system of Non linear Equations are as follows:
- Solve the following system of non linear equations:
y = x2
y = 8 – x2
y = 8 – x2
Solution:
y = x2
y = 8 – x2
y = x2
y = 8 – x2
y = y
x2 = 8 – x2
Both of these
sub-equations are true, but only the last one is usefully new and different:
x2 = 8 – x2
x2 = 8 – x2
2x2 = 8
x2 = 4
x = –2, +2
Then the solutions to
the original system will take place when x = –2 and when x = +2.
Plug up the x-values
into the first equation, since it's the simpler of the two:
x = –2:
y = x2
y = (–2)2 = 4
x = +2:
y = x2
y = (+2)2 = 4
Then the solutions are
(x, y) = (–2, 4) and (2, 4).
Example 2:
- Solve the following system of non linear equations:
b = a2
a2 + (b – 2)2 = 4
a2 + (b – 2)2 = 4
Solution:
b = a2
a2 + (b – 2)2 = 4
a2 + (b – 2)2 = 4
Substitute “b" for
"a2" in the second equation, and solve:
a2 + (b – 2)2 = 4
b + (b – 2)2 = 4
Therefore (b – 2)2 = (b2 – 4b + 4)
b + (b2 – 4b + 4) = 4
b2 – 3b = 0
b (b – 3) = 0
b = 0, b = 3
Locate the corresponding
x-values.
When b = 0:
b = a2
0 = a2
0 = a
0 = a2
0 = a
When b = 3:
b = a2
3 = a2
a = ±√ (3)
3 = a2
a = ±√ (3)
Answer: (0, 0), (-√3, 3) and (√3, 3)
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