Page 90 - Mathematics

Opening the brackets gives, Solution

15z − 10 = 30 Given x = 4.
15z = 40 Apply the definition of absolute
Dividing by 15 on both sides of the value. That is,
equation gives, ± x = 4
x
40 Either, x+ = 4 or − = 4
z =
15 Therefore, x = 4 or x = − 4.
8
=
3 Example 5.23
8
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Therefore, z = .
3 Solve for the value of x if 6 x− = 1.

Example 5.21 Solution
8 Given 6 x− = 1.
Solve for the value of y, if = 2.
3y − 2 Apply the definition of absolute
Solution value. That is, ( 6 x± − ) 1=

Given 8 = 2. Either,
3y − 2 + ( 6 x− ) 1 or= − ( 6 x− ) 1.=
Multiplying by (3y − 2) on both It follows that,
sides of the equation gives, 6 x− = 1 or − 6 x+ =
1
8 × (3y − 2) = 2(3y − 2) x= 5 or x = 7
(3y − 2) Therefore, x= 5 or x = .
7
8 2(3y= − 2)

86y= − 4 Example 5.24
12 = 6y
12 Solve for the value of x if
y = x + 2 = 2 and represent the
6
= 2 2. solution on a number line.
Therefore, y =
Mathematics Form One Example 5.22 4. Given x + ± ( x + + 2 = ) 2 = ) 2 = 2. Apply the ) 2 = 2.
Solution

definition of absolute value to get,
2

Find the values of x if x =
2 or −
( x +
Either ( x+

84

25/10/2024 09:51:27
Mathematics form 1.indd 84 25/10/2024 09:51:27
Mathematics form 1.indd 84