Question

$\frac{7}{x} + \frac{8}{y } = 2 \\ \frac{2}{x} + \frac{12}{y} = 20$
Can You Solve This Equation This is Class 10 Maths question..​

Answer 1

Answer:

${\boxed {\bf {\therefore x=-\dfrac{1}{2}\;and\;y=\dfrac{1}{2}}}}$

Step-by-step explanation:

Given:

$\longrightarrow \sf \dfrac{7}{x}+\dfrac{8}{y}=2\;\;\;\;\;..............(1)\\ \\ \\ \longrightarrow \sf \dfrac{2}{x}+\dfrac{12}{y}=20\;\;\;\;\;............(2)$

$\bf{Let,\;\dfrac{1}{x}=a\;and\;\dfrac{1}{y}=b,}$

$\longrightarrow \sf 7\Bigg(\dfrac{1}{x}\Bigg)+8\Bigg(\dfrac{1}{y}\Bigg)=2\\ \\ \\ \longrightarrow \sf 7a + 8b=2\;\;\;\;\;........(3) \\ \\ \rule{100}{1}\\ \\ \longrightarrow \sf 2\Bigg(\dfrac{1}{x}\Bigg)+12\Bigg(\dfrac{1}{y}\Bigg)=20\\ \\ \\ \longrightarrow \sf 2a + 12b=2\;\;\;\;\;........(4) \\ \\ \rule{100}{1}$

Now, solve equation (3) & (4) by substitution method,

$\longrightarrow \sf 7a+8b=2\;\;\;\;\;..........(3)\\ \\ \\ \longrightarrow \sf 7a=2-8b \\ \\ \\ \longrightarrow \sf a=\dfrac{2-8b}{7} \\ \\ \\ {\underline{\bf Now,\;put\;the\;value\;of\;`a'\;in\;equation\;(4),\;we\;get}}\\ \\ \\ \longrightarrow \sf 2a+12b=20\;\;\;\;\;.........(4)\\ \\ \\ \longrightarrow \sf 2\Bigg[\dfrac{2-8b}{7}\Bigg]+12b=20\\ \\ \\ \longrightarrow \sf \dfrac{4-16b}{7}+12b=20\\ \\ \\ \longrightarrow \sf \dfrac{4-16b+84b}{7}=20\\ \\ \\ \longrightarrow \sf 4+68b=140$

$\longrightarrow \sf 68b=140-4\\ \\ \\ \longrightarrow \sf 68b=136\\ \\ \\ \longrightarrow \sf b=\dfrac{136}{68}\\ \\ \\ \longrightarrow {\boxed{\bf b=2}}$

${\underline{\bf Put\;the\;value\;of\;`b'\;in\;equation\;(3),\;we\;get,}}$

$\longrightarrow \sf 7a+8b=2\\ \\ \\ \longrightarrow \sf 7a+8(2)=2\\ \\ \\ \longrightarrow \sf 7a + 16=2\\ \\ \\ \longrightarrow \sf 7a=2-16\\ \\ \\ \longrightarrow \sf 7a=-14\\ \\ \\ \longrightarrow \sf a=-\dfrac{14}{7}\\ \\ \\ \longrightarrow {\boxed{\bf a=-2}}$

$\sf Now, \\ \\ \\ \longrightarrow \sf \dfrac{1}{x}=a\\ \\ \\ \longrightarrow \sf \dfrac{1}{x}=-2\\ \\ \\ \longrightarrow {\boxed{\bf x=-\dfrac{1}{2}}} \\ \\ \rule{100}{1} \\ \\ \longrightarrow \sf \dfrac{1}{y}=b\\ \\ \\ \longrightarrow \sf \dfrac{1}{y}=2\\ \\ \\ \longrightarrow {\boxed{\bf y=\dfrac{1}{2}}}$

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