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Ask QuestionPosted by Piyush Daksha Prajapati Bsp 7 years, 9 months ago
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Posted by Akhil Rawat 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
If infinite number of solutions,
{tex}\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}{/tex}
or {tex}\frac{2}{{a + b + 1}} = \frac{3}{{a + 2b + 2}} = \frac{7}{{4a + 4b + 1}}{/tex}
If {tex}\frac{2}{{a + b + 1}} = \frac{3}{{a + 2b + 2}}{/tex}
{tex}\Rightarrow{/tex} a - b = 1
and if {tex}\frac{3}{{a + 2b + 2}} = \frac{7}{{4a + 4b + 1}}{/tex}
{tex}\Rightarrow{/tex} 5a - 2b = 11
On solving we get,
a = 3 and b = 2
Posted by Shiva Iyer 7 years, 9 months ago
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Posted by Tanvi Mansuri 7 years, 9 months ago
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Posted by Piyush Daksha Prajapati Bsp 7 years, 9 months ago
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Sia ? 6 years, 6 months ago
The given equations are
{tex}\frac { a } { x } - \frac { b } { y } = 0{/tex} ........... (i)
{tex}\frac { a b ^ { 2 } } { x } + \frac { a ^ { 2 } b } { y } = \left( a ^ { 2 } + b ^ { 2 } \right){/tex} ........... (ii)
Put {tex}\frac 1x{/tex}= u and {tex}\frac 1y{/tex}= v in the equation (i) and (ii), we get
{tex}\Rightarrow a u - b v = 0{/tex} .............(iii)
{tex}ab^2u + a^2bv - (a^2 + b^2) = 0{/tex}......(iv)
So, by cross multiplication,
{tex}\frac { u } { b _ { 1 } c _ { 2 } - b _ { 2 } c _ { 1 } } = \frac { v } { c _ { 1 } a _ { 2 } - c _ { 2 } a _ { 1 } } = \frac { 1 } { a _ { 1 } b _ { 2 } - a _ { 2 } b _ { 1 } }{/tex}
{tex}\Rightarrow \frac {u } { ( - b ) \left[ - \left( a ^ { 2 } + b ^ { 2 } \right) \right] - \left( a ^ { 2 } b \right) ( 0 ) }{/tex}{tex}= \frac { v } { ( 0 ) \left( a ^ { 2 } b \right) - \left[ - \left( a ^ { 2 } + b ^ { 2 } \right) \right] ( a ) } = \frac { 1 } { ( a ) \left( a ^ { 2 } b \right) - \left( a ^ { 2 } b \right) ( - b ) }{/tex}
{tex}\Rightarrow \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { v } { a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex} \frac { u } { b \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex}\Rightarrow \mathrm { u } = \frac { 1 } { \mathrm { a } }{/tex}
and {tex} \frac { v } { a \left( a ^ { 2 } + b ^ { 2 } \right) } = \frac { 1 } { a b \left( a ^ { 2 } + b ^ { 2 } \right) }{/tex}
{tex}\Rightarrow \mathrm { v } = \frac { 1 } { \mathrm { b } }{/tex}
If {tex} \mathrm { u } = \frac { 1 } { \mathrm { a } }{/tex}
{tex}\Rightarrow{/tex} {tex}x = a{/tex}
{tex} \mathrm { v } = \frac { 1 } { \mathrm { b } }{/tex}
{tex}\Rightarrow{/tex} {tex}y = b{/tex}
Posted by Devash Sharma 7 years, 9 months ago
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Harshita Vatyani 7 years, 9 months ago
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