CHAPTER 3 ADD MATH(3.1)

2020

CHAPTER 3 : SYSTEMS OF EQUATIONS
ADDITIONAL

MATHEMATICS
FORM 4
KSSM

SMK CONVENT SENTUL
3/21/2020

CHAPTER 3 : SYSTEMSS OF EQUATIONS

3.1 SYSTEMS OF LINEAR EQUATIONS IN THREE VARIABLES

TYPES OF SOLUTION

1. One solution – the planes intersect at only one point.

2. Infinite solutions - - the planes intersect along the straight line

3. No solution - the planes do not intersect at any point.

STEP TO SOLVE LINEAR EQUATIONS IN THREE VARIABLES.

STEP 1 : Write all the equations in standard form.
(cleared of decimals or fractions)

STEP 2 : Choose a variable to eliminate.
Then, form two linear equations with two variables.

[Use: Substitution / Elimination method]

STEP 3 : Solve the two equations from STEP 2 for two variables they contain.
[Use: Substitution / Elimination method]

STEP 4 : Substitute the solution back from STEP 3 into one of the original equations and solve for
the third variable.

STEP 1: STEP 2: STEP 3: STEP 4:
STANDARD FORM ELIMINATE 1 VARIABLE SOLVE EQUATIONS SUBSTITUTE 2 VARIABLES
FROM STEP 2 TO FIND
3 EQUATIONS FORM 2 EQUATIONS FROM STEP 3 INTO
3 VARIABLES WITH 2 VARIABLES 2 VARIABLES ORIGINAL EQUATION TO
FIND THIRD VARIABLES.
1 2 EQUATIONS

Example 1 :
Solve the following system of linear equations.

a) 3 − 2 + = 12
+ 3 + + 4 = 0 *not in standard form
2 + 2 − 4 = 6

Solution :
Step 1 : Write all in Standard form.

3 − 2 + = 12 → 3 − 2 + = 12 … … … … . ( )
+ 3 + + 4 = 0 → + 3 + = −4 … … … … . ( )
2 + 2 − 4 = 6 → 2 + 2 − 4 = 6 … … … … . ( )

Step 2 : Choose a variable to eliminate and form two linear equations in two variables.
[Use: Substitution / Elimination method] – eliminate variable of x

Eliminate (x): − Eliminate (x): −

3 − 2 + = 12 … … . . ( ) 2 + 6 + 2 = −8 … … . .2( )
(-) 3 + 9 + 3 = −12 … … 3( ) (-) 2 + 2 − 4 = 6 … … … … . ( )

−11y − 2z = 24 4y + 6z = −14

Step 3 : Solve the two linear equations in two variables in STEP 2.

−11y − 2z = 24 (×3)
4y + 6z = −14
→ ∶ −33y − 6z = 72
(eliminate : z) → ∶ (+) 4y + 6z = −14

−29y = 58
= −2

Substitute y=-2 into 4y + 6z = −14.

4(−2) + 6z = −14
−8 + 6z = −14

6z = −14 + 8
6z = −6

z = −1

STEP 4: Substitute the solution into the original equations and solve for third variable.

Substitute = −2 and z = −1 into ( ).
+ 3 + = −4 … … … … . ( )
+ 3(−2) + (−1) = −4
− 6 − 1 = −4
= −4 + 7
= 3

∴The solution is x=3, y=-2 and z=-1 (one solution)
2

b) + + = 2 … … … . ( ) All equations in standard form
2 + 2 + 2 = 4 ………(B)
−3 − 3 − 3 = −6 … . ( )

Solution:

Eliminate (x) : 2A-B

2 + 2 + 2 = 4 … … . .2
(-) 2 + 2 + 2 = 4 … … . .

0=0

∴The systems has infinite solution because 0=0.

c) − − 2 = 4 … … . ( ) → − − 2 = 4 … … … ( )

10( ) *cleared fractions
− − 0.8 = 0.2 … … ( ) → 4 − 4 − 8 = 2 … . ( ) or decimals.

− + + 2 = −3 …. ( ) → − + + 2 = −3 …. ( )

Solution :

Eliminate (x) : 4A-D

4 − 4 − 8 = 16
(-) 4 − 4 − 8 = 2

0 = 14

∴The system has no solution as ≠ .

*You do not have to complete any further steps.

CHECK YOUR ANSWER USING CALCULATOR:

3 − 2 + = 12 → a1=3, b1=-2, c1=1, d1=12 *equations must in
+ 3 + = −4 → a2=1, b2=3, c2=1, d2=-4 standard form : ax+by+cz=d
2 + 2 − 4 = 6 → a3=2, b3=2, c3=-4, d3=6

PRESS
MODE
EQN(1)→UNKNOWNS? (3)→a1? Press 3 = b1? Press -2= c1?press 1= d1? Press 12= a2?.......
until the screen show x=3, y=-2 and z=-1 .

PRACTICE : Textbook pg 76 (no. 1a, 2a)

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3.1.3 Solving problems involving systems of linear equations in three variables.
STEPS

• Understanding the problem
• Planning a strategy: determine the variable, form 3

equations
• Implementing the strategy: using 4 steps to solve equations
• Making a conclusion

EXAMPLE: Self Practice 3.3

1. Patricia invested RM24 500 in three unit trusts. She divided the money into three different unit
trust accounts, P, Q and R. At the end of the year, she obtained a profit of RM1 300. The annual
dividends for the accounts are 4%, 5.5% and 6% respectively. The final amount of money in account
P is four times that in account Q. How much money did she invest in each unit trust account?

Unit Trust P (p) Unit Trust Q (q) Unit Trust R (r) Total
1 1 1 24 500
4% 6% 1 300
1 5.5%
4q

Solution :

+ + = 24 500 → + + = 24 500 … … … … … … … ( )

(×1000 ) 40 + 55 + 60 = 1 300 000 … … ( )
4% + 5.5% + 6% = 1 300 →

= 4 → = 4 … … … . ( )

Eliminate a variable to form linear equations in two variables. [Use: Substitution method]

Substitute = /( ) into (A) Substitute = /( ) into (B)
+ + = 24 500 40 + 55 + 60 = 1 300 000
(4 ) + + = 24 500 40(4 ) + 55 + 60 = 1 300 000
5 + = 24 500 … … … . . ( ) 160 + 55 + 60 = 1 300 000
215 + 60 = 1 300 000 … … … . . ( )
5 + = 24 500 … … … . . ( )
215 + 60 = 1 300 000 … … … . . ( ) Two linear equations in two variables

From (D) : 5 + = 24 500 [Use: Substitution method] Substitute = into (F)
= 24 500 − 5 … … ( ) = 24 500 − 5 … … ( )
= 24 500 − 5(2000)
Substitute (F) into (E) : 215 + 60 = 1 300 000 = 14 500
215 + 60(24 500 − 5 ) = 1 300 000
215 + 1 470 000 − 300 = 1 300 000 Conclusion: = ,
−85 = 1 300 000 − 1 470 000 =
−85 = −170 000 = .
= 2000

Substitute q=2000 into (C): = 4
= 4(2000)
= 8000

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PRACTICE (Textbook pg 79 – no 2 and 3)

Instructions:

• Planning a strategy by complete the table below.

• Form systems of linear equations.

• Solve the systems of linear equations.

2.

Carnations(c) Roses(r) Daisies(d) Total

3. Pencil (y) Notebook (z) Total
Pen (x)

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