Answer the 2(ii) question."DO NOT SEND ANY LINKS".
Example 2. Solve the following system of simultaneous linear equations :
(ii) $$\begin{gathered} 4x+\frac{6}{y}=15 \\ 6x-\frac{8}{y}=14, y\ne 0 \end{gathered}$$

Dear Student,


$$\begin{gathered} To Solve : Following system of linear equations \\ 4x+\frac{6}{y}= 15 and \\ 6x-\frac{8}{y}=14, y\ne 0 \\ Solution\hspace{0.17em}: considering 4x+\frac{6}{y}=15 as equation \overline{)1} \\ and 6x-\frac{8}{y}=14 as equation \overline{)2} \\ Now, multiplying \overline{)1} with 4 and \overline{)2} with 3 and adding \\ \Rightarrow 4(4x+\frac{6}{y})+3(6x-\frac{8}{y})=(4\times 15)+(3\times 14) \\ \Rightarrow 16x+\frac{24}{y}+18x-\frac{24}{y}=60+42 \\ Rearranging the terms \\ \Rightarrow (16x+18x)+(\frac{24}{y}-\frac{24}{y})=102 \\ \Rightarrow 34x=102 \\ \Rightarrow x=\frac{102}{34} \\ \Rightarrow x=3 \\ Now, putting x=3 in \overline{)2} \\ \Rightarrow 6\times 3-\frac{8}{y}=14 \\ \Rightarrow 18-14=\frac{8}{y} \\ \Rightarrow 4=\frac{8}{y} \\ \Rightarrow y=\frac{8}{4} \\ \Rightarrow y=2 \\ Hence, for the given system of linear equations x will be 3 and y will be 2. \end{gathered}$$
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