Solve the following by substitution method …

{tex} Given\:pair\:of\: equations:\\\frac{x}{7}+\frac{y}{3}=5{/tex}

{tex}\implies \frac{y}{3}=5-\frac{x}{7}\:---(1){/tex}

{tex}and\:\frac{x}{2}-\frac{y}{9}=6{/tex}

{tex}\implies \frac{x}{2}-\frac{y}{3}\times \frac{1}{3}=6\:---(2){/tex}

{tex}substitute \:(1)\:in \: equation\:(2),\: we\:get{/tex}

{tex}\implies \frac{x}{2}-(5-\frac{x}{7})\times \frac{1}{3}=6\:---(2){/tex}

{tex}\implies \frac{x}{2}-\frac{5}{3}+\frac{x}{21}=6{/tex}

{tex}\implies \frac{x}{2}+\frac{x}{21}=6+\frac{5}{3}{/tex}

{tex}\implies \frac{21x+2x}{42}=\frac{18+5}{3}{/tex}

{tex}\implies \frac{23x}{42}=\frac{21}{3}{/tex}

{tex}\implies x = \frac{23}{3}\times \frac{42}{23}{/tex}

{tex}\implies x = 14\:--(3){/tex}

{tex}put \: x=14 \: in \: equation \:(1),\\we\:get{/tex}

{tex}\frac{y}{3}=5-\frac{14}{7}{/tex}

{tex}\implies \frac{y}{3}=5-2{/tex}

{tex}\implies y = 3\times 3 = 9{/tex}

Therefore,

{tex}x = 14 , \: y = 9{/tex}