4.1 Linear Equations I

Posted on July 1, 2017 by user

4.1 Linear Equations I

4.1.1 Equality

1. An equation is a mathematical statement that joins two equal quantities together by an equality sign ‘=’.

Example: 1 km = 1000 m

2. If two quantities are unequal, the symbol ‘≠’ (is not equal) is used.

Example: 9 ÷ 4 ≠ 3

4.1.2 Linear Equations in One Unknown

1. A linear algebraic term is a term with one unknown and the power of unknown is one.

Example: 8x, -7y, 0.5y, 3a, …..

2. A linear algebraic expression contains two or more linear algebraic terms which are joined by a plus or minus sign.

Example:

3x – 4y, 4x + 9, 6x – 2y + 5, ……

3. A linear equation is an equation involving numbers and linear algebraic terms.

Example:

5x – 4 = 11, 4x + 7 = 15, 3y – 2 = 7

4.1.3 Solutions of Linear Equations in One Unknown

1. Solving an equation is a process of finding the values of the unknown in the equation.

2. The number that satisfies the equation is called the solution or root of the equation.

Example 1:

x + 4 = 12

x = 12 – 4 ← (When +4 is moved to the right of the equation, it becomes –4)

x = 8

Example 2:

x – 7 = 11

x = 11 + 7 ← (When –7 is moved to the right of the equation, it becomes +7)

x = 18

Example 3:

\begin{array}{l} 8 x = 16 \\ x = \frac{16}{8} \leftarrow \begin{array}{l} when the multiplier 8 is moved \\ to the right of the equation, it \\ becomes the divisor 8 . \end{array} \\ x = 2 \end{array}

Example 4:

\begin{array}{l} \frac{x}{5} = 3 \\ x = 3 \times 5 \leftarrow \begin{array}{l} the divisor 5 becomes the \\ multiplier 5 when moved \\ to the right of the equation . \end{array} \\ x = 15 \end{array}

3.2.2 Algebraic Expressions II, PT3 Focus Practice

Posted on June 25, 2017 by user

Question 6:

Simplify - 4 (5 x - 3) + 7 x - 1.

Solution:

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

Question 7:

(3 x - 2 y) - (x + 4 y)

Solution:

\begin{array}{l} (3 x - 2 y) - (x + 4 y) \\ = 3 x - 2 y - x - 4 y \\ = 2 x - 6 y \end{array}

Question 8:

Simplify - 3 (2 a - 4 b) - \frac{1}{3} (6 a - 15 b)

Solution:

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

Question 9:

\frac{1}{3} (- 2 x + 6 y - 9 z) - \frac{1}{6} (- 4 x - 18 y + 24 z)

Solution:

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

Question 10:

\frac{1}{2} (a + 6 b c) - \frac{1}{5} (3 + 2 b c - 2 a)

Solution:

\begin{array}{l} \frac{1}{2} (a + 6 b c) - \frac{1}{5} (3 + 2 b c - 2 a) \\ = \frac{1}{2} a + 3 b c - \frac{3}{5} - \frac{2}{5} b c + \frac{2}{5} a \\ = \frac{5 a + 4 a}{10} + \frac{15 b c - 2 b c}{5} - \frac{3}{5} \\ = \frac{9 a}{10} + \frac{13 b c}{5} - \frac{3}{5} \end{array}

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}

3.1 Algebraic Expressions II

Posted on June 20, 2017 by user

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

(c) 20 x y \times (- \frac{2}{5} x^{2} y)

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} 20 x y \times (- \frac{2}{5} x^{2} y) \\ = 20^{4} x y \times (- \frac{2}{5}) \times x \times x \times 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} (a) \frac{42 x y z}{7 x y} \\ (b) \frac{12 x y^{2}}{18 x y} \end{array}

(c) 35 p^{2} q r^{2} \div 30 p r

Solution:

\begin{array}{l} (a) \frac{42 x y z}{7 x y} \\ = \frac{42^{6} x \times y \times z}{7 x \times y} = 6 z \end{array}

\begin{array}{l} (b) \frac{12 x y^{2}}{18 x y} \\ = \frac{12^{2} x \times y \times y}{18^{3} x \times y} = \frac{2}{3} y \end{array}

\begin{array}{l} (c) 35 p^{2} q r^{2} \div 30 p r \\ = \frac{35^{7} p \times p \times q \times r \times r}{30^{6} p \times r} \\ = \frac{7}{6} p q r \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

Posted on June 17, 2017 by user

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

Solution:

2.2.3 Squares, Square Roots, Cube and Cube Roots, PT3 Practice

Posted on June 16, 2017 by user

Question 10:

\begin{array}{l} (a) Find the value of \sqrt{0.09} \\ (b) Calculate the value of {(\frac{5}{7} \times \sqrt{49})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt{0.09} = \sqrt{\frac{9}{100}} = \frac{3}{10} = 0.3 \\ (b) \\ {(\frac{5}{7} \times \sqrt{49})}^{2} = {(\frac{5}{7} \times 7)}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 11:

\begin{array}{l} (a) Find the value of \sqrt[3]{- \frac{1}{27}} \\ (b) Calculate the value of {(7 + \sqrt[3]{- 8})}^{2} \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{- \frac{1}{27}} = \sqrt[3]{- \frac{1}{3} \times - \frac{1}{3} \times - \frac{1}{3}} = - \frac{1}{3} \\ (b) \\ {(7 + \sqrt[3]{- 8})}^{2} = {[7 + (- 2)]}^{2} \\ = 5^{2} \\ = 25 \end{array}

Question 12:

\begin{array}{l} (a) Find the value of {(- \frac{1}{4})}^{3} . \\ (b) Calculate the value of {(4.2 \div \sqrt[3]{27})}^{2} . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{1}{4})}^{3} = - \frac{1}{4} \times - \frac{1}{4} \times - \frac{1}{4} \\ = - \frac{1}{64} \\ (b) \\ {(4.2 \div \sqrt[3]{27})}^{2} = {(4.2 \div 3)}^{2} \\ = {1.4}^{2} \\ = 1.4 \times 1.4 \\ = 1.96 \end{array}

Question 13:

\begin{array}{l} (a) Find the value of \sqrt[3]{0.008} . \\ (b) Calculate the value of 3^{2} \times \sqrt[3]{- \frac{64}{27}} . \end{array}

Solution:

\begin{array}{l} (a) \\ \sqrt[3]{0.008} = \sqrt[3]{\frac{8}{1000}} \\ = \frac{2}{10} \\ = 0.2 \\ (b) \\ 3^{2} \times \sqrt[3]{- \frac{64}{27}} = 9^{3} \times - \frac{4}{3} \\ = 12 \end{array}

Question 14:

\begin{array}{l} (a) Find the value of {(- \frac{3}{4})}^{3} . \\ (b) Calculate the value of \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) . \end{array}

Solution:

\begin{array}{l} (a) \\ {(- \frac{3}{4})}^{3} = - \frac{3}{4} \times - \frac{3}{4} \times - \frac{3}{4} \\ = - \frac{27}{64} \\ (b) \\ \sqrt[3]{2 \frac{10}{27}} \div (2^{2} - 3^{2}) = \sqrt[3]{\frac{64}{27}} \div (4 - 9) \\ = \frac{4}{3} \div - 5 \\ = \frac{4}{3} \times - \frac{1}{5} \\ = - \frac{4}{15} \end{array}

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

Posted on June 14, 2017 by user

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.

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{}{3} \div \frac{5}{4} \\ = \frac{}{3} \times \frac{4}{5} \\ = \end{array}

Solution:

\begin{array}{l} \sqrt[3]{4 \frac{17}{27}} \div \sqrt{1 \frac{9}{16}} = \sqrt[3]{\frac{125}{27}} \div \sqrt{\frac{25}{16}} \\ = \frac{5}{3} \div \frac{5}{4} \\ = \frac{5}{3} \times \frac{4}{5} \\ = \frac{4}{3} \end{array}

Question 7:

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

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{}{9}} - \frac{3}{})}^{2} \\ = {(\frac{}{12})}^{2} \\ = \end{array}

Solution:

\begin{array}{l} {(\sqrt{2 \frac{7}{9}} - \sqrt[3]{\frac{27}{64}})}^{2} = {(\sqrt{\frac{25}{9}} - \frac{3}{4})}^{2} \\ = {(\frac{5}{3} - \frac{3}{4})}^{2} \\ = {(\frac{(5 \times 4) - (3 \times 3)}{12})}^{2} \\ = {(\frac{11}{12})}^{2} \\ = \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{}{64}} - {0.3}^{2} \\ = \frac{}{4} - \\ = \end{array}

Solution:

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

Question 9:

Find the value of:

\begin{array}{l} (a) \sqrt[3]{\frac{10}{27} - 5} \\ (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} \end{array}

Solution:

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

\begin{array}{l} (b) \sqrt{\frac{4}{49}} \times \sqrt[3]{- 0.216} = \frac{2}{7} \times \sqrt[3]{\frac{216}{1000}} \\ = \frac{2}{7} \times \frac{6}{510} \\ = \frac{6}{35} \end{array}

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

Posted on June 12, 2017 by user

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

Question 1:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt{\frac{50}{98}} \\ (b) \sqrt{1 \frac{17}{64}} \\ (c) \sqrt{81} - \sqrt{0.01} \\ (d) \sqrt{3.24} \end{array}

Solution:

(a) \sqrt{\frac{50}{98}} = \sqrt{\frac{5025}{9849}} = \sqrt{\frac{25}{49}} = \frac{5}{7}

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

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

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

Question 2:

Calculate the values of the following:

\begin{array}{l} (a) \sqrt[3]{\frac{16}{250}} \\ (b) \sqrt[3]{- \frac{4}{256}} \\ (c) \sqrt[3]{0.008} \\ (d) \sqrt[3]{0.729} \end{array}

Solution:

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

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

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

\begin{array}{l} (d) \sqrt[3]{- 0.729} = \sqrt[3]{- \frac{729}{1000}} \\ = - \frac{9}{10} \\ = - 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}} \\ = \frac{3}{2} + \frac{8}{5} \\ = \frac{31}{10} = 3 \frac{1}{10} \end{array}

Question 4:

Find the values of the following:

(a) 1 – (–0.3)³.

(b)

{(2.1 \div \sqrt[3]{27})}^{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} {(2.1 \div \sqrt[3]{27})}^{2} = {(2.1 \div 3)}^{2} \\ = {(0.7)}^{2} \\ = 0.49 \end{array}

Question 5:

Find the values of the following:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} \\ (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \end{array}

Solution:

\begin{array}{l} (a) {(9 + \sqrt[3]{- 8})}^{2} = {[9 + (- 2)]}^{2} \\ = 7^{2} \\ = 49 \end{array}

\begin{array}{l} (b) \sqrt{144} \div \sqrt[3]{216} \times {0.3}^{3} \\ = 144 \div 6 \times (0.3 \times 0.3 \times 0.3) \\ = 24 \times 0.027 \\ = 0.648 \end{array}

2.1 Squares, Square Roots, Cube and Cube Roots

Posted on June 10, 2017 by user

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) 13²= 13 × 13 = 169

(b) (–10)²= (–10) × (–10) = 100

(d) (–0.06)²= (–0.06) × (–0.06) = 0.0036

\begin{array}{l} (e) {(3 \frac{1}{2})}^{2} = {(\frac{7}{2})}^{2} = \frac{7}{2} \times \frac{7}{2} = \frac{49}{4} \\ (f) {(- 1 \frac{2}{7})}^{2} = {(- \frac{9}{7})}^{2} = (- \frac{9}{7}) \times (- \frac{9}{7}) = \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} (a) \sqrt{169} = \sqrt{13 \times 13} = 13 \\ (b) \sqrt{\frac{25}{64}} = \sqrt{\frac{5 \times 5}{8 \times 8}} = \frac{5}{8} \\ (c) \sqrt{\frac{72}{98}} = \sqrt{\frac{7236}{9849}} = \sqrt{\frac{6 \times 6}{7 \times 7}} = \frac{6}{7} \\ (d) \sqrt{3 \frac{1}{16}} = \sqrt{\frac{49}{16}} = \frac{7}{4} = 1 \frac{3}{4} \\ (e) \sqrt{1.44} = \sqrt{1 \frac{4411}{10025}} = \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 \times 4 \times 4} \\ = 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 \times 5 \times 5} \\ = 5 \end{array}

3. The cube root of a negative number is negative.

Example:

\begin{array}{l} \sqrt[3]{- 125} = \sqrt[3]{(- 5) \times (- 5) \times (- 5)} \\ = - 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{168}{250125}} \\ = \sqrt[3]{\frac{8}{125}} \\ = \frac{2}{5} \end{array}

1.2.2 Directed Numbers, PT3 Practice

Posted on June 7, 2017 by user

1.2.1 Directed Numbers, PT3 Practice 2

Question 6:

Calculate the value of

1 - (- 2 + 7 \times 0.15) \div 2 \frac{1}{2}

Solution:

\begin{array}{l} 1 - (- 2 + 7 \times 0.15) \div 2 \frac{1}{2} \\ = 1 - (- 2 + 1.05) \times \frac{2}{5} \\ = 1 - (- 0.95) \times \frac{2}{5} \\ = 1 - (- 0.38) \\ = 1.38 \end{array}

Question 7:

Calculate the value of

1 \frac{1}{8} \times (\frac{4}{5} - \frac{2}{3})

and express the answer as a fraction in its lowest terms.

Solution:

\begin{array}{l} 1 \frac{1}{8} \times (\frac{4}{5} - \frac{2}{3}) \\ = 1 \frac{1}{8} \times (\frac{12}{15} - \frac{10}{15}) \\ = \frac{9^{3}}{8_{4}} \times \frac{2^{1}}{15_{5}} \\ = \frac{3}{20} \end{array}