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System of Non-Linear Equations Start this lesson by answering the warm-up questions below. This is to assess the ideas or concepts related to Non-linear and Linear equations.
Activity 1.5.a Linear Against Non-Linear Equations Answer the following questions below:
1. Determine if the equation is linear or not
a. 2x + 3y – 4 = 0
b. xy + 2 = 0
c. x = y
d. y = 0
e. x = 2y + 3 2. Which of the following is a non-linear equation?
a. 2x2 + 3y – 4 = 0
b. xy + 2 = 0
c. x2 – y = 2
d. 2x + 3y – 6 = 0
e. x = y
System of Non-Linear Equations
I. Definition of System of Non-Linear Equations
Non-linear equations involve two or more equations of the second degree in the form Ax2 + Bxy + Cy2 + Dx + Ey + F = 0. Non-linear curves are circle, parabola, ellipse, and hyperbola. Possible non-linear equations involve the combinations of the equations of the circle, parabola, ellipse, and hyperbola.*…show more content…*

The solution of the system is the intersection point of the graphs of the given equations. II. Solution of the System A system of equations can be solved through Eliminations method, Substitution method or by a Graphical method. A. By elimination method Example 1: Solve the system of equation by elimination method x2 + y2 = 4 - eq’n (1) x2 - y2 = 4 - eq’n (2) Assigning equation (1) & (2) and eliminate y variable by adding the two equations, we have 2x2 = 8, and solving for x x2 = 4 → x = ±2 Using the value of x and solving for y using eq’n (1) (±2)2 + y2 = 4 4 + y2 = 4 y2 = 4 - 4 y2 = 0 → y =

The solution of the system is the intersection point of the graphs of the given equations. II. Solution of the System A system of equations can be solved through Eliminations method, Substitution method or by a Graphical method. A. By elimination method Example 1: Solve the system of equation by elimination method x2 + y2 = 4 - eq’n (1) x2 - y2 = 4 - eq’n (2) Assigning equation (1) & (2) and eliminate y variable by adding the two equations, we have 2x2 = 8, and solving for x x2 = 4 → x = ±2 Using the value of x and solving for y using eq’n (1) (±2)2 + y2 = 4 4 + y2 = 4 y2 = 4 - 4 y2 = 0 → y =

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