# 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

# 2.2.4 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Question 15:
Match each of the following with the correct value.

Solution:

Question 10:

Solution:

Question 11:

Solution:

Question 12:

Solution:

Question 13:

Solution:

Question 14:

Solution:

# 2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

2.2.2 Squares, Square Roots, Cube and Cube Roots, PT3 Focus Practice

Question 6:
Complete the operation steps below by filling in the boxes using suitable numbers.

Solution:

$\begin{array}{l}\sqrt[3]{4\frac{17}{27}}÷\sqrt{1\frac{9}{16}}=\sqrt[3]{\frac{\overline{)125}}{27}}÷\sqrt{\frac{25}{16}}\\ \text{}=\frac{5}{3}÷\frac{5}{4}\\ \text{}=\frac{\overline{)5}}{3}×\frac{4}{\overline{)5}}\\ \text{}=\overline{)\frac{4}{3}}\end{array}$

Question 7:
Complete the operation steps below by filling in the boxes using suitable numbers.
$\begin{array}{l}{\left(\sqrt{2\frac{7}{9}}-\sqrt[3]{\frac{27}{64}}\right)}^{2}={\left(\sqrt{\frac{\overline{)}}{9}}-\frac{3}{\overline{)}}\right)}^{2}\\ \text{}={\left(\frac{\overline{)}}{12}\right)}^{2}\\ \text{}=\overline{)}\end{array}$

Solution:
$\begin{array}{l}{\left(\sqrt{2\frac{7}{9}}-\sqrt[3]{\frac{27}{64}}\right)}^{2}={\left(\sqrt{\frac{\overline{)25}}{9}}-\frac{3}{\overline{)4}}\right)}^{2}\\ \text{}={\left(\frac{5}{3}-\frac{3}{4}\right)}^{2}\\ \text{}={\left(\frac{\left(5×4\right)-\left(3×3\right)}{12}\right)}^{2}\\ \text{}={\left(\frac{\overline{)11}}{12}\right)}^{2}\\ \text{}=\overline{)\frac{121}{144}}\end{array}$

Question 8:
Complete the operation steps below by filling in the boxes using suitable numbers.
$\begin{array}{l}\sqrt[3]{1\frac{61}{64}}-{0.3}^{2}=\sqrt[3]{\frac{\overline{)}}{64}}-{0.3}^{2}\\ \text{}=\frac{\overline{)}}{4}-\overline{)}\\ \text{}=\overline{)}\end{array}$

Solution:
$\begin{array}{l}\sqrt[3]{1\frac{61}{64}}-{0.3}^{2}=\sqrt[3]{\frac{\overline{)125}}{64}}-{0.3}^{2}\\ \text{}=\frac{\overline{)5}}{4}-\overline{)0.09}\\ \text{}=1.25-0.09\\ \text{}=\overline{)1.16}\end{array}$

Question 9:
Find the value of:
$\begin{array}{l}\text{(a)}\sqrt[3]{\frac{10}{27}-5}\\ \text{(b)}\sqrt{\frac{4}{49}}×\sqrt[3]{-0.216}\end{array}$

Solution:
$\begin{array}{l}\text{(a)}\sqrt[3]{\frac{10}{27}-5}=\sqrt[3]{\frac{10-135}{27}}\\ \text{}=\sqrt[3]{-\frac{125}{27}}\\ \text{}=-\frac{5}{3}\end{array}$

$\begin{array}{l}\text{(b)}\sqrt{\frac{4}{49}}×\sqrt[3]{-0.216}=\frac{2}{7}×\sqrt[3]{\frac{216}{1000}}\\ \text{}=\frac{\overline{)2}}{7}×\frac{6}{5\overline{)10}}\\ \text{}=\frac{6}{35}\end{array}$

# 2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

2.2.1 Squares, Square Roots, Cube and Cube Roots, PT3 Practice 1

Question 1:
Calculate the values of the following:
$\begin{array}{l}\text{(a)}\sqrt{\frac{50}{98}}\\ \text{(b)}\sqrt{1\frac{17}{64}}\\ \text{(c)}\sqrt{81}-\sqrt{0.01}\\ \text{(d)}\sqrt{3.24}\end{array}$

Solution:
$\text{(a)}\sqrt{\frac{50}{98}}=\sqrt{\frac{\overline{)50}25}{\overline{)98}49}}=\sqrt{\frac{25}{49}}=\frac{5}{7}$

$\text{(b)}\sqrt{1\frac{17}{64}}=\sqrt{\frac{81}{64}}=\frac{9}{8}=1\frac{1}{8}$

$\begin{array}{l}\text{(c)}\sqrt{81}-\sqrt{0.01}=9-\sqrt{\frac{1}{100}}\\ \text{}=9-\frac{1}{10}\\ \text{}=9-0.1\\ \text{}=8.9\end{array}$

$\begin{array}{l}\text{(d)}\sqrt{3.24}=\sqrt{3\frac{24}{100}}=\sqrt{3\frac{6}{25}}\\ \text{}=\sqrt{\frac{81}{25}}\\ \text{}=\frac{9}{5}=1\frac{4}{5}\end{array}$

Question 2:
Calculate the values of the following:
$\begin{array}{l}\text{(a)}\sqrt[3]{\frac{16}{250}}\\ \text{(b)}\sqrt[3]{-\frac{4}{256}}\\ \text{(c)}\sqrt[3]{0.008}\\ \text{(d)}\sqrt[3]{0.729}\end{array}$

Solution:
$\text{(a)}\sqrt[3]{\frac{16}{250}}=\sqrt[3]{\frac{8}{125}}=\frac{2}{5}$

$\text{(b)}\sqrt[3]{-\frac{4}{256}}=\sqrt[3]{-\frac{1}{64}}=-\frac{1}{4}$

$\begin{array}{l}\text{(c)}\sqrt[3]{0.008}=\sqrt[3]{\frac{8}{1000}}\\ \text{=}\frac{2}{10}\\ \text{}=0.2\end{array}$

$\begin{array}{l}\text{(d)}\sqrt[3]{-0.729}=\sqrt[3]{-\frac{729}{1000}}\\ \text{}=-\frac{9}{10}\\ \text{}=-0.9\end{array}$

Question 3:
Find the value of $\sqrt[3]{3\frac{3}{8}}+\sqrt{2\frac{14}{25}}.$

Solution:
$\begin{array}{l}\sqrt[3]{3\frac{3}{8}}+\sqrt{2\frac{14}{25}}=\sqrt[3]{\frac{27}{8}}+\sqrt{\frac{64}{25}}\\ \text{}=\frac{3}{2}+\frac{8}{5}\\ \text{}=\frac{31}{10}=3\frac{1}{10}\end{array}$

Question 4:
Find the values of the following:
(a) 1 – (–0.3)3.
(b) ${\left(2.1÷\sqrt[3]{27}\right)}^{2}$

Solution:
(a)
1 – (–0.3)3 = 1 – [(–0.3) × (–0.3) × (–0.3)]
= 1 – (–0.027)
= 1 + 0.027
= 1.027

(b)
$\begin{array}{l}{\left(2.1÷\sqrt[3]{27}\right)}^{2}={\left(2.1÷3\right)}^{2}\\ \text{}={\left(0.7\right)}^{2}\\ \text{}=0.49\end{array}$

Question 5:
Find the values of the following:
$\begin{array}{l}\text{(a)}{\left(9+\sqrt[3]{-8}\right)}^{2}\\ \text{(b)}\sqrt{144}÷\sqrt[3]{216}×{0.3}^{3}\end{array}$

Solution:
$\begin{array}{l}\text{(a)}{\left(9+\sqrt[3]{-8}\right)}^{2}={\left[9+\left(-2\right)\right]}^{2}\\ \text{}={7}^{2}\\ \text{}=49\end{array}$

$\begin{array}{l}\text{(b)}\sqrt{144}÷\sqrt[3]{216}×{0.3}^{3}\\ \text{}=144÷6×\left(0.3×0.3×0.3\right)\\ \text{}=24×0.027\\ \text{}=0.648\end{array}$

# 2.1 Squares, Square Roots, Cube and Cube Roots

2.1 Squares, Square Roots, Cube and Cube Roots

(A) Squares
The square of a number is the answer you get when you multiply a number by itself.

Example:
(a) 13= 13 × 13 = 169
(b)   (–10)= (–10) × (–10) = 100
(c) (0.4)2 = 0.4 × 0.4 = 0.16
(d)   (–0.06)= (–0.06) × (–0.06) = 0.0036
$\begin{array}{l}\text{(e)}{\left(3\frac{1}{2}\right)}^{2}={\left(\frac{7}{2}\right)}^{2}=\frac{7}{2}×\frac{7}{2}=\frac{49}{4}\\ \left(\text{f}\right)\text{}{\left(-1\frac{2}{7}\right)}^{2}={\left(-\frac{9}{7}\right)}^{2}=\left(-\frac{9}{7}\right)×\left(-\frac{9}{7}\right)=\frac{81}{49}\end{array}$

(B) Perfect Squares
1. Perfect squares are the squares of whole numbers.

2. Perfect squares are formed by multiplying a whole number by itself.
Example:
4 = 2 × 2   9 = 3 × 3   16 = 4 × 4

3. The first twelve perfect squares are:
= 12, 22, 32, 42, 52, 62, 72, 82, 92, 102, 112, 122
= 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144

(C) Square Roots
1. The square root of a positive number is a number multiplied by itself whose product is equal to the given number.

Example:
$\begin{array}{l}\text{(a)}\sqrt{169}=\sqrt{13×13}=13\\ \text{(b)}\sqrt{\frac{25}{64}}=\sqrt{\frac{5×5}{8×8}}=\frac{5}{8}\\ \text{(c)}\sqrt{\frac{72}{98}}=\sqrt{\frac{\overline{)72}36}{\overline{)98}49}}=\sqrt{\frac{6×6}{7×7}}=\frac{6}{7}\\ \text{(d)}\sqrt{3\frac{1}{16}}=\sqrt{\frac{49}{16}}=\frac{7}{4}=1\frac{3}{4}\\ \text{(e)}\sqrt{1.44}=\sqrt{1\frac{\overline{)44}11}{\overline{)100}25}}=\sqrt{\frac{36}{25}}=\frac{6}{5}=1\frac{1}{5}\end{array}$

(D) Cubes
1. The cube of a number is obtained when that number is multiplied by itself twice.
Example:
The cube of 3 is written as
33 = 3 × 3 × 3
= 27

2.
The cube of a negative number is negative.
Example:
(–2)3 = (–2) × (–2) × (–2)
= –8
3. The cube of zero is zero. The cube of one is one, 13 = 1.

(E) Cube Roots
1. The cube root of a number is a number which, when multiplied by itself twice, produces the particular number. $"\sqrt[3]{}"$  is the symbol for cube root.
Example:
$\begin{array}{l}\sqrt[3]{64}=\sqrt[3]{4×4×4}\\ \text{}=4\end{array}$
$\sqrt[3]{64}$ is read as ‘cube root of sixty-four’.

2.
The cube root of a positive number is positive.
Example:
$\begin{array}{l}\sqrt[3]{125}=\sqrt[3]{5×5×5}\\ \text{}=5\end{array}$

3.
The cube root of a negative number is negative.
Example:
$\begin{array}{l}\sqrt[3]{-125}=\sqrt[3]{\left(-5\right)×\left(-5\right)×\left(-5\right)}\\ \text{}=-5\end{array}$

4.
To determine the cube roots of fractions, the fractions should be simplified to numerators and denominators that are cubes of integers.
Example:
$\begin{array}{l}\sqrt[3]{\frac{16}{250}}=\sqrt[3]{\frac{\overline{)16}8}{\overline{)250}125}}\\ \text{}=\sqrt[3]{\frac{8}{125}}\\ \text{}=\frac{2}{5}\end{array}$

# 1.2.2 Directed Numbers, PT3 Practice

1.2.1 Directed Numbers, PT3 Practice 2
Question 6:
Calculate the value of
$1-\left(-2+7×0.15\right)÷2\frac{1}{2}$

Solution:
$\begin{array}{l}1-\left(-2+7×0.15\right)÷2\frac{1}{2}\\ =1-\left(-2+1.05\right)×\frac{2}{5}\\ =1-\left(-0.95\right)×\frac{2}{5}\\ =1-\left(-0.38\right)\\ =1.38\end{array}$

Question 7:
Calculate the value of   $1\frac{1}{8}×\left(\frac{4}{5}-\frac{2}{3}\right)$  and express the answer as a fraction in its lowest terms.

Solution:
$\begin{array}{l}1\frac{1}{8}×\left(\frac{4}{5}-\frac{2}{3}\right)\\ =1\frac{1}{8}×\left(\frac{12}{15}-\frac{10}{15}\right)\\ =\frac{{\overline{)9}}^{3}}{{\overline{)8}}_{4}}×\frac{{\overline{)2}}^{1}}{{\overline{)15}}_{5}}\\ =\frac{3}{20}\end{array}$

# 1.2.1 Directed Numbers, PT3 Practice

1.2.1 Directed Numbers, PT3 Practice 1
Question 1:
Complete the following calculation.
$\begin{array}{l}-4.6×\left(\frac{3}{5}-1\frac{3}{4}\right)=-4.6×\left(0.6-\overline{)x}\right)\\ \text{}=-4.6×\overline{)x}\\ \text{}=\overline{)x}\end{array}$

Solution:
$\begin{array}{l}-4.6×\left(\frac{3}{5}-1\frac{3}{4}\right)=-4.6×\left(0.6-\overline{)1.75}\right)\\ \text{}=-4.6×\overline{)-1.15}\\ \text{}=\overline{)5.29}\end{array}$

Question 2:
Complete the following calculation.
$\begin{array}{l}-43+35÷1\frac{1}{6}=-43+35÷\frac{7}{6}\\ \text{}=-43+35×\overline{)x}\\ \text{}=-43+\overline{)x}\\ \text{}=-13\end{array}$

Solution:
$\begin{array}{l}-43+35÷1\frac{1}{6}=-43+35÷\frac{7}{6}\\ \text{}=-43+35×\overline{)\frac{6}{7}}\\ \text{}=-43+\overline{)30}\\ \text{}=-13\end{array}$

Question 3:
Calculate each of the following.
(a) (–9) × 14
(b)   6.7 – 3.2 × (–0.5)
(c) –144 ÷ (–8) ÷ (–2)

Solution:
(a) (–9) × 14 = –126
(b)   6.7 – 3.2 × (–0.5) = 6.7 – (–1.6)
= 6.7 + 1.6
= 8.3
(c) –144 ÷ (–8) ÷ (–2) = –144 ÷ 4
= 36

Question 4:
Calculate the value of
$0.6+\left(-1\frac{2}{3}\right)÷\left(\frac{4}{15}\right)-2$

Solution:
$\begin{array}{l}0.6+\left(-1\frac{2}{3}\right)÷\left(\frac{4}{15}\right)-2\\ =0.6+\left[-\frac{5}{\overline{)3}}×\frac{{\overline{)15}}^{5}}{4}\right]-2\\ =0.6+\left[-\frac{25}{4}\right]-2\\ =0.6-6\frac{1}{4}-2\\ =0.6-6.25-2\\ =-7.65\end{array}$

Question 5:
Calculate the value of
$\left(-0.6×1\frac{1}{4}\right)-\left[-1.6×\left(-1.5\right)\right]÷0.4$

Solution:
$\begin{array}{l}\left(-0.6×1\frac{1}{4}\right)-\left[-1.6×\left(-1.5\right)\right]÷0.4\\ =\left({\overline{)-0.6}}^{-0.3}×\frac{5}{{\overline{)4}}^{2}}\right)-\left[\frac{{\overline{)-1.6}}^{-4}×\left(-1.5\right)}{{\overline{)0.4}}^{1}}\right]\\ =-\frac{1.5}{2}-6\\ =-0.75-6\\ =-6.75\end{array}$