Question 2.30: Simplify the following algebraic expressions: 1. 3ab + 2ac −...

1.       All that is required here is to add or subtract like terms, so we get:

3ab + 5ab − 4ab + 2ac − 2ac − 3c + 2c − b

= 4ab − b − c.

2.       Here you need to be aware of the law of precedence. This is derived from the laws of arithmetic you learned earlier. As an aide-mémoire we use the acronym BODMAS: brackets, of, division, multiplication, addition and finally subtraction, with these operations being performed in this order. From this law we carry out multiplication before addition or subtraction. So we get:

3x − 8yz − 2x = x − 8yz.

3.      With this expression, when multiplying the brackets, we need to remember the law of indices for multiplication. Using this law, we get:

6a^{2−1}b^{2−1}c^{2−1} + 4a^{1−1}b^{1−1}c^{1−1}\\  = 6a^1 b^1 c^1 + 4a^0 b^0 c^0 = 6abc + 4.

(Do not forget the 4! Remember that any number raised to the power zero is 1 and 4 × 1 × 1 × 1 = 4.)

4.      This is just the multiplication of brackets where we multiply all terms in the right-hand bracket by both terms in the left-hand bracket. We perform these multiplications as though we are reading a book from left to right. Starting with (3x) × (2x) = 6x², then (3x) × (−3y) = −9xy and so on. We then repeat the multiplications, using the right-hand term in the first bracket: (2y) × (2x) = 4xy and so on. So, before any simplification, we should end up with 2 × 3 = 6 terms:

(3x + 2y)(2x − 3y + 6z)

= 6x² − 9xy + 18xz + 4xy

− 6y² + 12yz.

After simplification, which involves only two like terms in this case, we get:

6x² − 5xy + 18xz − 6y² − 12yz.