\[ \large\begin{eqnarray} 3 \times 4 - 2 \times 5 & = & \hbox{............ (1)}\\ & & \\ (3 \times 4) - (2 \times 5) & = & \hbox{............ (2)}\\ \left((3 \times 4) - 2\right) \times 5 & = & \hbox{............ (3)}\\ 3 \times (4 - 2) \times 5 & = & \hbox{............ (4)}\\ 3 \times \left(4 - (2 \times 5)\right) & = & \hbox{............ (5)}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} 10 - 2^2 + 3 &=& \hbox{........... (1)}\\ & & \\ {3^2 \times 7 - 3 + 6 \over 4 + 2} &=& \hbox{........... (2)}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} 10^2 \times 10^3 & = & 10^5\\ 100 \times 1000 & = & 100,000\\ & & \\ {10^{5}\over 10^2} = 10^5 \times 10^{-2} & = & 10^{(5-2)} = 10^3\\ & & \\ \left(10^3\right)^2 = 10^3 \times 10^3 & = & 10^{3+3} = 10^6 \\ \end{eqnarray} \]

\[ \large\begin{eqnarray} 10^2 \times 10^3 \times 10^4 & = & 10^{(2 + 3 + 4)} = 10^7\\ \\ a^3 \times a^2 \times a & = & a^{(3 + 2 + 1)} = a^6\\ \\ a^x \times a^y \times a^z & = & a^{(x+y+z)}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} {a^{7} \over a^{2}} = a^7 \times a^{-2} & = & a^{(7-2)} = a^5\\ \\ {a^{x} \over a^{y}} = a^x \times a^{-y} & = & a^{(x-y)}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} {1\over 2} + {1\over 4} - {3\over 8} = {4 \over 8} + {2\over 8} - {3\over 8} & = & {4 + 2 -3 \over 8} = {3\over 8}\\ & & \\ 1{1\over 3} + 4{1\over6} = {3 + 1\over 3}+ {24+1\over 6} = {4\over 3} + {25\over 6} & = & {8 + 25\over 6} = {33\over 6} = {11 \over 2}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} {1\over 2} \times {2\over 5} \times {3\over 8} & = & {1 \times 2 \times 3 \over 2 \times 5 \times 8} = {6 \over 80} = {3\over 40}\\ & & \\ {2{1 \over 3} \over 4{2\over 5}} = {{7\over 3}\over {22\over 5}} & = & {7 \times 5\over 22 \times 3} = {35\over 66}\\ \end{eqnarray} \]

\[ \large\begin{eqnarray} 2{2\over 3} \times {1\over 5} + 1{1\over 2} \div 2{1\over 2} & = & \hbox{.......... (1)}\\ & & \\ {{\left({1\over 3} \right)^2}\times 4{1\over 2}\over 3} & = & \hbox{.......... (2)}\\ \end{eqnarray} \]

Identity for an operation is the quantity that will not change the value of number

\[ \large 3 \times 1 = 3 \]

Give an example and determine the value of \( \large I \) for each relationship.

\[ \large\begin{array}{rlcrl} a \times I = a & \hbox{.......... (1)} & & {a \over I} = a & \hbox{.......... (2)}\\ a - I = a & \hbox{.......... (3)} & & a + I = a & \hbox{.......... (4)}\\ a^I = a & \hbox{.......... (5)} & & a \times a^I = a & \hbox{.......... (6)}\\ \end{array} \]

A quantity that will cause an operation result in the identity value.

\[ \large 4 \times {1\over 4} = 1 \]

Identify the reciprocal for each relationship.

\[ \large\begin{array}{rlcrl} a \times I = a & \hbox{.......... (1)} & & {a \over I} = a & \hbox{.......... (2)}\\ a - I = a & \hbox{.......... (3)} & & a + I = a & \hbox{.......... (4)}\\ a^I = a & \hbox{.......... (5)} & & a \times a^I = a & \hbox{.......... (6)}\\ \end{array} \]

\[ \large\begin{eqnarray} (a + b) + c & = & a + (b + c)\\ (1 + 3) + 4 & = & 1 + (3 + 4)\\ 4 + 4 & = & 1 + 7\\ & & \\ (ab)c &=& a(bc)\\ (2\times 3) \times 5 & = & 2 \times (3 \times 5)\\ 6 \times 5 & = & 2 \times 15\\ \end{eqnarray} \]

• What is the cost per kilo? ………… (1)

• If the market value is US$25.00 per 500 gm, what is the profit margin? ………… (2)

• Basic algebraic concepts
• Simplification of fractions
• Implication of equations
• Solving equations