JSS 1 Mathematics with Assignment

Lesson Two

Class: Jss1

Subject: Mathematics

Topic: Simple Linear Equations

Sub– Topic: Equation involving Brackets

Equations involving brackets are equations that contain brackets.

Steps to solve equations with brackets are as follows:

1. A. Open the brackets by:

(I) Multiplying the number in front of the bracket by all the numbers inside the bracket.

Note: If a negative number is in front of a bracket, all the signs inside the brackets change on opening the bracket, but if a positive number is in front of a bracket, all the signs inside the bracket remain the same on opening the bracket.

1. B. Collect like terms
2. C. Divide both sides by the coefficient to get the unknown.

 

Example 1: Solve the following equations.

(1). 3(a – 2) + 2a = 24  (2).  4(y + 3) = 6(y – 1)  (3).  5(f + 3) – 2(f – 1) = 26

 

         Solution

1. 3(a-2) + 2a =24

    Open bracket

      3a – 6 + 2a = 24

    Collect like terms

      3a + 2a = 24 + 6

      5a = 30

    Divide both sides by 5

       5a/5 = 30/5

       a = 6

 

2.  4(y+3)=6(y-1)

    Open brackets

       4y + 12= 6y – 6

    Collect like terms

       12 + 6 =6y – 4y

        18 = 2y

    Divide both sides by 2

        18/2 = 2y/2

        9 = y

        y = 9

 

3. 5(f + 3) – 2(f – 1) = 26

    Open brackets

      5f + 15 – 2f + 2 = 26  (Note when a negative number is in front of the bracket)

    Collect like terms

      5f – 2f = 26 – 15 – 2

      3f = 9

    Divide both sides by 3

      3f/3 = 9/3

       f = 3

 

             ASSIGNMENT

 

    Solve the following equations

1. 4(y – 6) – 3y = 25

 

2. 5(r + 3) – 2(r – 4) = 53

 

3. 3(k + 1) + 2(k – 3) = 17

 

Download here  JSS1 Mathematics Lesson Two

 

To be submitted on Friday 22nd May, 2020 via whatsApp @ 07035773326.

N/B. For further studies, refer to New Concept Mathematics for JSS1 page 166 – 177