To solve equations like 2a + 5 = 3a - 6, it shouldn't be too tough.
Your mission is to solve for the unknown a in the case of the above equation.
To solve for the unknown , a, you have to move all the unknown terms(2a and 3a) to one side and all constant terms (5 and -6) to the other side.
2a + 5 = 3a - 6
I will move the smaller unknown term, 2a to the R.H.S of the equation. In order to do that, I will subtract L.H.S by 2a (-2a). To make the equation balance, I need to subtract R.H.S by 2a too.
2a + 5 - 2a = 3a - 6 - 2a
So on the L.H.S, '2a' will be gone as (2a- 2a = 0) and on R.H.S 3a- 2a = a.
5 = a - 6
Now, you will move the constant terms to the L.H.S. You will add 6 to the R.H.S. To balance the equation, you will need to add 6 to the L.H.S.
I am sure you have gone through Lesson #1 on Solving of Equations Involving addition, subtraction, multiplication and division.
You can go through your textbook of Unit 2.2.1 and 2.2.2 from pg 36 to 39 for better understanding.Remember to do the practice questions from worksheet 29.
Here's a video with 4 more examples on previous lesson.
Examples of Solving Equations Involving 1 Step
Here's Lesson #2 -> Solving Equations Involving More Than One Operation You may refer to Unit 2.2.3 (pg 39 - pg 41) of your textbook. At the end of the four video lessons, Attempt the following questions from textbook, Exercise 2D pg 41. 1(a) ,1(c), 1(g), 1(h), 2(b), 2(c), 2(k), 2(h) 3(a), 3(b), 3(c), 3(d)
Solve Equation of the form Ax + B = C
More Examples of Solving Equation of the Form Ax + B = C
Solve Equation of the form A(Bx + C) = D
More Examples of Solving Equation of the Form A(Bx + C) = D
At the end of the four video lessons, Attempt the following questions from textbook, Exercise 2D pg 41. 1(a) ,1(c), 1(g), 1(h), 2(b), 2(c), 2(k), 2(h) 3(a), 3(b), 3(c), 3(d)
