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Linear Equations (I)

4.2.2 Linear Equations I, PT3 Practice

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Question 6:
Solve each of the following equations:
(a)  5 a − 16 = a (b)  6 + 2 3 ( − 9 p + 12 ) = p


Solution:
(a)
5 a − 16 = a 5 a − a = 16 4 a = 16 a = 4

(b)
6 + 2 3 ( − 9 p + 12 ) = p 6 − 6 p + 8 = p − 6 p − p = − 8 − 6 − 7 p = − 14 p = 2


Question 7::
Solve each of the following equations:
(a)  a − 2 = a 5 (b)  b − 3 4 = 2 + b 5

Solution:
(a)
a − 2 = a 5 5 a − 2 = a 4 a = 2 a = 2

(b)
b − 3 4 = 2 + b 5 5 b − 15 = 8 + 4 b 5 b − 4 b = 8 + 15 b = 23

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Question 8:
Solve each of the following equations:
(a)  x = − 24 − x (b)  y + 5 4 ( 4 − 2 y ) = − 4

Solution:
(a)
x = − 24 − x 2 x = − 24 x = − 12

(b)
y + 5 4 ( 4 − 2 y ) = − 4 4 y + 20 − 10 y = − 16    ( × 4 ) − 6 y = − 16 − 20 − 6 y = − 36 y = 6


Question 9:
Solve each of the following equations:
(a)  5 p = 8 p − 9 (b)  3 q = 20 − 13 q 4

Solution:
(a)
5 p = 8 p − 9 5 p − 8 p = − 9 − 3 p = − 9 p = 3

(b)
3 q = 20 − 13 q 4 12 q = 20 − 13 q 12 q + 13 q = 20 25 q = 20 q = 4 5

4.2.1 Linear Equations I, PT3 Practice 1

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4.2.1 Linear Equations I, PT3 Practice 1
Question 1:
Solve the following linear equations.
  (a)  4 – 3n = 5n – 4   
  (b) 4 m − 2 10 + m = 1 2  

Solution:
(a)
4 – 3n = 5n – 4   
–3n –5n = – 4 – 4   
–8n = – 8   
 8n = 8
n = 8/8 = 1
 
(b) 4 m − 2 10 + m = 1 2 2 ( 4 m − 2 ) = 10 + m 8 m − 4 = 10 + m 8 m − m = 10 + 4 7 m = 14 m = 14 7 m = 2


Question 2:
Solve the following linear equations.
(a) x 3 = 4 − x (b) 3 ( x − 2 ) 5 = 9
 
Solution:
(a) x 3 = 4 − x x = 3 ( 4 − x ) x = 12 − 3 x x + 3 x = 12 4 x = 12 x = 12 4 x = 3

(b) 3 ( x − 2 ) 5 = 9 3 ( x − 2 ) = 9 × 5 3 x − 6 = 45 3 x = 45 + 6 3 x = 51 x = 51 3 x = 17



Question 3:
Solve the following linear equations.
(a) 3 m 4 + 15 = 9 (b) 2 m + 8 = 3 ( m − 2 )

Solution:
(a) 3 m 4 + 15 = 9 3 m 4 = 9 − 15 3 m 4 = − 6 3 m = − 6 × 4 3 m = − 24 m = − 24 3 m = − 8
 
(b) 2 m + 8 = 3 ( m − 2 ) 2 m + 8 = 3 m − 6 2 m − 3 m = − 6 − 8 − m = − 14 m = 14


Question 4:
Solve the following linear equations.
(a) 11 + 2 x 3 = 9 (b) x − 5 3 = x 6

Solution:
(a) 11 + 2 x 3 = 9 2 x 3 = 9 − 11 2 x 3 = − 2 2 x = − 6 x = − 6 2 x = − 3
 
(b) x − 5 3 = x 6 6 ( x − 5 ) = 3 x 6 x − 30 = 3 x 6 x − 3 x = 30 3 x = 30 x = 10


Question 5:
Solve the following linear equations.
(a) 4 ( 2 x − 3 ) = 24 (b) y 2 − y + 4 3 = 5

Solution:
(a) 4 ( 2 x − 3 ) = 24 8 x − 12 = 24 8 x = 24 + 12 8 x = 36 x = 36 8 x = 4 1 2
 
(b) y 2 − y + 4 3 = 5 y × 3 2 × 3 − 2 ( y + 4 ) 3 × 2 = 5 3 y − 2 ( y + 4 ) 6 = 5 3 y − 2 y − 8 = 30 y = 30 + 8 y = 38

4.1 Linear Equations I

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4.1 Linear Equations I
 
4.1.1 Equality
1. An equation is a mathematical statement that joins two equal quantities together by an equality sign ‘=’.
Example: 1 km = 1000 m

2. If two quantities are unequal, the symbol ‘≠’ (is not equal) is used.
Example: 9 ÷ 4 ≠ 3


4.1.2 Linear Equations in One Unknown
1. A linear algebraic term is a term with one unknown and the power of unknown is one.
Example: 8x, -7y, 0.5y, 3a, …..

2. A linear algebraic expression contains two or more linear algebraic terms which are joined by a plus or minus sign.
Example:
3x – 4y, 4x + 9, 6x – 2y + 5, ……

3. A linear equation is an equation involving numbers and linear algebraic terms.
Example:
5x – 4 = 11, 4x + 7 = 15, 3y – 2 = 7


4.1.3 Solutions of Linear Equations in One Unknown
1. Solving an equation is a process of finding the values of the unknown in the equation.
2. The number that satisfies the equation is called the solution or root of the equation.
Example 1:
x + 4 = 12
  x = 12 – 4 ← (When +4 is moved to the right of the equation, it becomes –4)
  x = 8

Example 2:
x – 7 = 11
  x = 11 + 7 ← (When –7 is moved to the right of the equation, it becomes +7)
  x = 18

Example 3:
8 x = 16 x = 16 8 ← when the multiplier 8 is moved to the right of the equation, it becomes the divisor 8 . x = 2

Example 4:
x 5 = 3 x = 3 × 5 ← the divisor 5 becomes the multiplier 5 when moved to the right of the equation . x = 15