# My Add Maths Modules - Form 4 - Simultaneous Equations - With Past Year SPM Questions

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```					                  MY
Mathematics
Modules
Form 4
(Version 2011)

Topic 4
MASTER
in

by

NgKL
(M.Ed.,B.Sc.Hons.Dip.Ed.,Dip.Edu.Mgt.,Cert.NPQH.)
TOPIC 4 – SIMULTANEOUS EQUATIONS
4.1   IMPORTANT NOTES:

4.1.1 Simultaneous Equations In Two Unknowns (Variables):
One is linear equation and the other one is non-linear equation.
4.1.2 Linear Equation is an algebraic equation, in which the variable (unknown) or variables (unknowns) are
of the first degree (that is, the power to its variable or variables is one).
Example: x + 2y = 4; 2x − 5y = 7; x = 7; y = − 3

Other equations which do not have the characteristics as mentioned for a                     linear equation are
categorised as Non-Linear Equations.
1 1
Example: x2 + y2 – 5x + 3y + 8 = 0; xy + 3y = 2;               2 , y = 2x2 – 5; etc.
x y
To solve these kind of simultaneous equations is by Substitution Method.
4.1.3 Steps to solve the simultaneous equations;
1.   Identify the linear equation.
Example:       x + 2y = 1 ..........(1)                  This is the linear equation.
2x2 + y2 = 7 ..........(2)
2.   Express one of the unknowns in the linear equation in terms of the other unknown.
Example:        x = 1 – 2y ......... (3)     or     2y = 1 – x
1 x
y=        ..........(3)
2
3.   Substitute the unknown expressed in Step 2 into the non-linear equation to obtain a quadratic
equation of the other unknown.
Example: (3)           (2)     2(1 – 2y)2 + y2 = 4
2(1 – 4y + 4y2 ) + y2 = 7
2 – 8y + 8y2 + y2 – 7 = 0
9y2 – 8y – 5 = 0             A quadratic equation
4.   Solve the quadratic equation to obtain the values of the unknown by one of the three methods;
(a) Factorisation, (x + a)(x + b) = 0; only if the equation can be factorised.

(b) Completing square, (Studied in Topic 2, Quadratic Equation).

 b  b 2  4ac
(c) Quadratic formula, x or y 
2a
Refer to the quadratic equation obtained in Step 3, list the values of a, b and c.
a = 9, b = − 8, c = −5                  substitute these values into the formula.

2
 ( 8 )  ( 8 )  4( 9 )( 5 )
y   
2( 9 )
8  244
y=             ;   y = 1.312 ( in 4 s.f)
18
8  64  180                                  8  244
y                                                y=             ;   y = − 7.620 (in 4 s.f)
18                                          18

5.   Substitute each of the y values obtained into equation (3) to find the each of the x values;
y = 1.312; x = 1 – 2(1.312) = −1.64 #
y = −7.620; x = 1 – 2(−7.62) = 16.2 4#

2
Exercise 4.1

1. Identify whether each of the following equations is a linear equation or a non-linear equation.

x       y
1.         4x + y = 8                                     6.                   1
3       5
2       6
2.         (x – 2)2 + 8 = 0                               7.                   1
x       y

3.         x+y–9=0                                        8.            x + 2y – 15 = 0

4.         x2 – 10 = 2y                                   9.    2y + 9x = 7

2y
5.         3(2 + x2) + xy = 0                             10.           3y  3
x

2. State the respective value of a, b and c in the following quadratic equations.

No.                         Quadratic Equation                      a                      b       c

1.         x2 + 3x + 6 = 0

2.         5y2 – 11y + 8 = 0

3.         3p2 + 4p – 3 = 10

4.         2x2 – px + 2x + p + 7 = 0

4.2        SOLVING SIMULTANEOUS EQUATIONS
(a)     Solve the following simultaneous equations
3x – y + 6 = 0
2x2 + y2 = 11

3
(b)   Solve the following simultaneous equations
x+ y=7
xy = 10

2    2
(c)   Solve the following simultaneous equations: 3x + 2y = x + y + xy = 7

Note:

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