Answer:
[tex]\qquad \:\boxed{\begin{aligned}& \qquad \:\sf \:(a) \: - \dfrac{3}{5} \times ( - 1) = \dfrac{3}{5} \qquad \: \\ \\& \qquad \:\sf \: (b) \: - \dfrac{5}{7} \times \dfrac{7}{5} = - 1\end{aligned}} \qquad \\ \\ [/tex]
Step-by-step explanation:
[tex]\large\underline{\sf{Solution-a}}[/tex]
Given expression is
[tex]\sf \: - \dfrac{3}{5} \times ...... = \dfrac{3}{5} \\ \\ [/tex]
Let assume that missing number be x.
So,
[tex]\sf \: - \dfrac{3}{5} \times x = \dfrac{3}{5} \\ \\ [/tex]
[tex]\sf \: x = \dfrac{3}{5} \div \bigg( - \dfrac{3}{5} \bigg) \\ \\ [/tex]
[tex]\sf \: x = - \: \dfrac{3}{5} \div \bigg( \dfrac{3}{5} \bigg) \\ \\ [/tex]
[tex]\sf \: x = - \: \dfrac{3}{5} \times \dfrac{5}{3} \\ \\ [/tex]
[tex]\sf\implies \sf \: x = - \: 1 \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: - \dfrac{3}{5} \times ( - 1) = \dfrac{3}{5} \\ \\ [/tex]
[tex]\large\underline{\sf{Solution-b}}[/tex]
[tex]\sf \: - \dfrac{5}{7} \times ...... = - 1 \\ \\ [/tex]
Let missing number be x
[tex]\sf \: - \dfrac{5}{7} \times x = - 1 \\ \\ [/tex]
[tex]\sf \: x = - 1 \div \bigg( - \dfrac{5}{7} \bigg) \\ \\ [/tex]
[tex]\sf \: x = 1 \div \bigg(\dfrac{5}{7} \bigg) \\ \\ [/tex]
[tex]\sf \: x = 1 \times \dfrac{7}{5} \\ \\ [/tex]
[tex]\sf\implies \sf \: x = \dfrac{7}{5} \\ \\ [/tex]
Hence,
[tex]\sf\implies \bf \: - \dfrac{5}{7} \times \dfrac{7}{5} = - 1 \\ \\ [/tex]