3.2.1 Algebraic Expressions II, PT3 Focus Practice

Posted on June 23, 2017 by user

3.2.1 Algebraic Expressions II, PT3 Focus Practice

Question 1:

Calculate the product of each of the following pairs of algebraic terms.

(a) 2rs × 4r²s³t

$\begin{array}{l} (b) 1 \frac{2}{5} a^{2} b^{2} \times \frac{5}{14} a b^{3} c^{2} \\ (c) 1 \frac{1}{2} x y \times \frac{4}{9} x^{2} z \end{array}$

Solution:

(a)

2rs × 4r²s³t = 2 × r × s× 4 × r × r × s × s × s × t = 8 r³s⁴t

$\begin{array}{l} (b) 1 \frac{2}{5} a^{2} b^{2} \times \frac{5}{14} a b^{3} c^{2} \\ = \frac{7^{1}}{5^{1}} a a b b \times \frac{5^{1}}{14^{2}} a b b b c c \\ = \frac{1}{2} a^{3} b^{5} c^{2} \end{array}$

$\begin{array}{l} (c) 1 \frac{1}{2} x y \times \frac{4}{9} x^{2} z \\ = \frac{3^{1}}{2^{1}} x y \times \frac{4^{2}}{9^{3}} x x z \\ = \frac{2}{3} x^{3} y z \end{array}$

Question 2:

Find the quotients of each of the following pairs of algebraic terms.

$(a) \frac{- 48 a^{2} b^{3} c^{4}}{- 16 a b^{2} c^{2}}$

$\begin{array}{l} (b) - 15 e f^{3} g^{2} \div (- 40 e f^{2} g) \\ (c) 5 a^{3} c^{2} \div \frac{1}{2} a c \end{array}$

Solution:

$\begin{array}{l} (a) \frac{- 48 a^{2} b^{3} c^{4}}{- 16 a b^{2} c^{2}} \\ = \frac{- 48^{3} a \times a \times b \times b \times b \times c \times c \times c \times c}{- 16^{1} a \times b \times b \times c \times c} \\ = 3 a c^{2} \end{array}$

$\begin{array}{l} (b) - 15 e f^{3} g^{2} \div (- 40 e f^{2} g) \\ = \frac{- 15^{3} e \times f \times f \times f \times g \times g}{- 40^{8} e \times f \times f \times g} \\ = \frac{3}{8} f g \end{array}$

$\begin{array}{l} (c) 5 a^{3} c^{2} \div \frac{1}{2} a c = 5 a {^{3}}^{2} c^{2} \times \frac{2}{a c} \\ = 10 a^{2} c \end{array}$

Question 3:

(– 4a²b) ÷ 3b²c²× 9ac =

Solution:

$\begin{array}{l} (- 4 a^{2} b) \div 3 b^{2} c^{2} \times 9 a c \\ = \frac{- 4 a^{2} b}{3 b^{21} c^{21}} \times 9^{3} a c \\ = \frac{- 12 a^{3}}{b c} \end{array}$

Question 4:

2a²b × 14b³c ÷ 56ab²c² =

Solution:

$\begin{array}{l} 2 a^{2} b \times 14 b^{3} c \div 56 a b^{2} c^{2} \\ = 2 a^{21} b \times \frac{14 b^{31} c}{56^{2} a b^{2} c^{21}} \\ = \frac{a b^{2}}{2 c} \end{array}$

Question 5:

$- \frac{2}{3} (3 a - 6 b + \frac{3}{4} c) =$

Solution:

$\begin{array}{l} - \frac{2}{3} (3 a - 6 b + \frac{3}{4} c) \\ = - \frac{2}{3} \times 3 a - \frac{2}{3} (- 6^{2} b) - \frac{2}{3} (\frac{3}{4^{2}} c) \\ = - 2 a + 4 b - \frac{1}{2} c \end{array}$