**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:

**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.

**B**.**Collect****like****terms****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**

**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**

** **

**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**

** **

**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**

**4(y – 6) – 3y = 25**

** **

**5(r + 3) – 2(r – 4) = 53**

** **

**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 **

Please, what of J’s 2 Mathematics, I can’t find it.

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