# 3.2.2 Algebraic Expressions II, PT3 Focus Practice

Question 6:

Solution:
$\begin{array}{l}-4\left(5x-3\right)+7x-1\\ =-20x+12+7x-1\\ =-13x+11\end{array}$

Question 7:
$\left(3x-2y\right)-\left(x+4y\right)$

Solution:
$\begin{array}{l}\left(3x-2y\right)-\left(x+4y\right)\\ =3x-2y-x-4y\\ =2x-6y\end{array}$

Question 8:

Solution:
$\begin{array}{l}-3\left(2a-4b\right)-\frac{1}{3}\left(6a-15b\right)\\ =-6a+12b-2a+5b\\ =-8a+17b\end{array}$

Question 9:
$\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)$

Solution:
$\begin{array}{l}\frac{1}{3}\left(-2x+6y-9z\right)-\frac{1}{6}\left(-4x-18y+24z\right)\\ =\overline{)-\frac{2}{3}x}+2y-3z\overline{)+\frac{2}{3}x}+3y-4z\\ =5y-7z\end{array}$

Question 10:
$\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)$

Solution:
$\begin{array}{l}\frac{1}{2}\left(a+6bc\right)-\frac{1}{5}\left(3+2bc-2a\right)\\ =\frac{1}{2}a+3bc-\frac{3}{5}-\frac{2}{5}bc+\frac{2}{5}a\\ =\frac{5a+4a}{10}+\frac{15bc-2bc}{5}-\frac{3}{5}\\ =\frac{9a}{10}+\frac{13bc}{5}-\frac{3}{5}\end{array}$

# 3.2.1 Algebraic Expressions II, PT3 Focus Practice

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

Solution:
(a)
2rs × 4r2s3t = 2 × r × s× 4 × r × r × s × s × s × t = 8 r3s4t

$\begin{array}{l}\text{(b)}1\frac{2}{5}{a}^{2}{b}^{2}×\frac{5}{14}a{b}^{3}{c}^{2}\\ =\frac{{\overline{)7}}^{1}}{{\overline{)5}}^{1}}aabb×\frac{{\overline{)5}}^{1}}{{\overline{)14}}^{2}}abbbcc\\ =\frac{1}{2}{a}^{3}{b}^{5}{c}^{2}\end{array}$

$\begin{array}{l}\text{(c)}1\frac{1}{2}xy×\frac{4}{9}{x}^{2}z\\ =\frac{{\overline{)3}}^{1}}{{\overline{)2}}^{1}}xy×\frac{{\overline{)4}}^{2}}{{\overline{)9}}^{3}}xxz\\ =\frac{2}{3}{x}^{3}yz\end{array}$

Question 2:
Find the quotients of each of the following pairs of algebraic terms.
$\text{(a)}\frac{-48{a}^{2}{b}^{3}{c}^{4}}{-16a{b}^{2}{c}^{2}}$
$\begin{array}{l}\text{(b)}-15e{f}^{3}{g}^{2}÷\left(-40e{f}^{2}g\right)\\ \text{(c)}5{a}^{3}{c}^{2}÷\frac{1}{2}ac\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\frac{-48{a}^{2}{b}^{3}{c}^{4}}{-16a{b}^{2}{c}^{2}}\\ =\frac{-{\overline{)48}}^{3}\overline{)a}×a×\overline{)b}×\overline{)b}×b×\overline{)c}×\overline{)c}×c×c}{-{\overline{)16}}^{1}\overline{)a}×\overline{)b}×\overline{)b}×\overline{)c}×\overline{)c}}\\ =3a{c}^{2}\end{array}$
$\begin{array}{l}\text{(b)}-15e{f}^{3}{g}^{2}÷\left(-40e{f}^{2}g\right)\\ =\frac{-{\overline{)15}}^{3}\overline{)e}×\overline{)f}×\overline{)f}×f×\overline{)g}×g}{-{\overline{)40}}^{8}\overline{)e}×\overline{)f}×\overline{)\overline{)f}}×\overline{)g}}\\ =\frac{3}{8}fg\end{array}$

$\begin{array}{l}\text{(c)}5{a}^{3}{c}^{2}÷\frac{1}{2}ac=5a{\overline{){}^{3}}}^{2}c\overline{){}^{2}}×\frac{2}{\overline{)a}\overline{)c}}\\ \text{}=10{a}^{2}c\end{array}$

Question 3:
(– 4a2b) ÷ 3b2c2× 9ac =

Solution:
$\begin{array}{l}\left(-4{a}^{2}b\right)÷3{b}^{2}{c}^{2}×9ac\\ =\frac{-4{a}^{2}\overline{)b}}{\overline{)3}{b}^{\overline{)2}1}{c}^{\overline{)2}1}}×{\overline{)9}}^{3}a\overline{)c}\\ =\frac{-12{a}^{3}}{bc}\end{array}$

Question 4:
2a2b × 14b3c ÷ 56ab2c2 =

Solution:
$\begin{array}{l}2{a}^{2}b×14{b}^{3}c÷56a{b}^{2}{c}^{2}\\ =\overline{)2}{a}^{\overline{)2}1}b×\frac{\overline{)14}{b}^{\overline{)3}1}\overline{)c}}{{\overline{)56}}^{2}\overline{)a}{\overline{)b}}^{\overline{)2}}{c}^{\overline{)2}1}}\\ =\frac{a{b}^{2}}{2c}\end{array}$

Question 5:
$-\frac{2}{3}\left(3a-6b+\frac{3}{4}c\right)=$

Solution:
$\begin{array}{l}-\frac{2}{3}\left(3a-6b+\frac{3}{4}c\right)\\ =-\frac{2}{\overline{)3}}×\overline{)3}a-\frac{2}{\overline{)3}}\left(-{\overline{)6}}^{2}b\right)-\frac{\overline{)2}}{\overline{)3}}\left(\frac{\overline{)3}}{{\overline{)4}}^{2}}c\right)\\ =-2a+4b-\frac{1}{2}c\end{array}$

# 3.1 Algebraic Expressions II

3.1 Algebraic Expressions II

3.1.1 Algebraic terms in Two or More Unknowns
1. An algebraic term in two or more unknowns is the product of the unknowns with a number.
Example 1:
4a3b = 4 × a × a × a× b

2.
The coefficient of an unknown in the given algebraic term is the factor of the unknown.
Example 2:
7ab: Coefficient of ab is 7.

3. Like algebraic terms
are algebraic terms with the same unknowns.

3.1.2 Multiplication and Division of Two or More Algebraic Terms
The coefficients and the unknowns of algebraic terms can be multiplied or divided altogether.

Example 3:
Calculate the product of each of the following pairs of algebraic terms.
(a) 5ac × 2bc
(b) –6xy × 5yz
$\text{(c)}20xy×\left(-\frac{2}{5}{x}^{2}y\right)$

Solution:
(a)
5ac × 2bc = 5 × a × c × 2 × b × c = 10abc 2

(b)
–6xy × 5yz = –6 × x × y × 5 × y × z = –30 xy2z

(c)
$\begin{array}{l}20xy×\left(-\frac{2}{5}{x}^{2}y\right)\\ ={\overline{)20}}^{4}xy×\left(-\frac{2}{\overline{)5}}\right)×x×x×y\\ =-8{x}^{3}{y}^{2}\end{array}$

Example 4:
Find the quotients of each of the following pairs of algebraic terms.
$\begin{array}{l}\text{(a)}\frac{42xyz}{7xy}\\ \text{(b)}\frac{12x{y}^{2}}{18xy}\end{array}$
$\text{(c)}35{p}^{2}q{r}^{2}÷30pr$

Solution:
$\begin{array}{l}\text{(a)}\frac{42xyz}{7xy}\\ =\frac{{\overline{)42}}^{6}\overline{)x}×\overline{)y}×z}{\overline{)7}\overline{)x}×\overline{)y}}=6z\end{array}$
$\begin{array}{l}\text{(b)}\frac{12x{y}^{2}}{18xy}\\ =\frac{{\overline{)12}}^{2}\overline{)x}×\overline{)y}×y}{{\overline{)18}}^{3}\overline{)x}×\overline{)y}}=\frac{2}{3}y\end{array}$
$\begin{array}{l}\text{(c)}35{p}^{2}q{r}^{2}÷30pr\\ =\frac{{\overline{)35}}^{7}\overline{)p}×p×q×\overline{)r}×r}{{\overline{)30}}^{6}\overline{)p}×\overline{)r}}\\ =\frac{7}{6}pqr\end{array}$

3.1.3 Algebraic Expressions
An algebraic expression contains one or more algebraic terms. These terms are separated by a plus or minus sign.

Example 5:
7 – 6a2 b + c is an algebraic expression with 3 terms.

3.1.4 Computation Involving Algebraic Expressions
Computation Involving Algebraic Expressions:
(a)   2(3a – 4) = 6a – 8
(b)   (15a – 9b) ÷ 3 = 5a – 3b
(c) (6a – 2) – (9 + 4a)
= 6a – 4a – 2 – 9
= 2a – 11
(d)   (a2 b – 5ab2) – (6a2 b – 4abc – 6ab2)
= a2 b – 6a2 b – 5ab2 – (– 6ab2) – (– 4abc)
= –5a2 b + ab2 + 4abc