Math, asked by amritjha68, 8 months ago

solve the following pair of linear equations by cross-multiplication method: 8x+2y=9
3x+2y=4.

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Answers

Answered by Anonymous
16

S O L U T I O N :

\bf{\large{\underline{\bf{Given\::}}}}}

  • 8x + 2y = 9
  • 3x + 2y = 4

\bf{\large{\underline{\bf{Explanation\::}}}}}

We have;

\mapsto\sf{8x+2y=9}\\\\\mapsto\sf{8x+2y-9=0....................(1)}

&

\mapsto\sf{3x+2y=4}\\\\\mapsto\sf{3x+2y-4=0....................(2)}

So;

\bullet\:\tt{a_1=8}\\\bullet\tt{a_2=3}\\\bullet\tt{b_1=2}\\\bullet\tt{b_2=2}\\\bullet\tt{c_1=-9}\\\bullet\tt{c_2=-4}

\bigstar By using cross - multiplication :

\mapsto\tt{\bigg[\dfrac{x}{b_1c_2-b_2c_1} =\dfrac{y}{c_1a_2-c_2a_1} =\dfrac{1}{a_1b_2-a_2b_1} \bigg]}\\\\\\\mapsto\tt{\dfrac{x}{2\times (-4)-2\times (-9)} =\dfrac{y}{-9\times 3-(-4)\times 8} =\dfrac{1}{8\times 2-3\times 2} }\\\\\\\mapsto\tt{\dfrac{x}{-8-(-18)} =\dfrac{y}{-27-(-32)} =\dfrac{1}{16-6} }\\\\\\\mapsto\tt{\dfrac{x}{-8+18} =\dfrac{y}{-27+32} =\dfrac{1}{10} }\\\\\\\mapsto\tt{\dfrac{x}{10} =\dfrac{y}{5} =\dfrac{1}{10} }\\\\\\

\mapsto\tt{\dfrac{x}{10} =\dfrac{1}{10} \:\:\:Or\:\:\:\dfrac{y}{5} =\dfrac{1}{10} }\\\\\\\mapsto\tt{10x=10\:\:\:Or\:\:\:10y=5}\\\\\\\mapsto\tt{x=\cancel{10/10} \:\:\:Or\:\:\:y=\cancel{5/10}}\\\\\\\mapsto\bf{x=1\:\:\:Or\:\:\:y=1/2}

Thus;

The value of x and y will be 1 & 1/2 .

Answered by Anonymous
18

{\huge{\bf{\red{\underline{Solution:}}}}}

{\bf{\blue{\underline{Given:}}}}

  \dagger \:  \: {\sf{ { \green{8x + 2y = 9}}}} \\

  \dagger \:  \: {\sf{ { \green{3x + 2y = 4}}}} \\

{\bf{\blue{\underline{Now:}}}}

 : \implies{\sf{  \frac{x}{ \frac{2}{2}    \times \frac{9}{4}   } =  \frac{y}{ \frac{9}{4}  \times  \frac{8}{3} } =  \frac{ - 1}{ \frac{8}{3}  \times  \frac{2}{2} }  }} \\ \\

 : \implies{\sf{  \frac{x}{(4)(2) - (2)(9)}  =  \frac{y}{(9)(3) - (8)(4)}  =  \frac{ - 1}{(8)(2) - (3)(2)} }} \\ \\

 : \implies{\sf{  \frac{x}{8 - 18}  =  \frac{y}{27 - 32}  =  \frac{ - 1}{16 - 6} }} \\ \\

 : \implies{\sf{  \frac{x}{ - 10}  =  \frac{y}{ - 5}  =  \frac{ - 1}{10} }} \\ \\

Now,

 : \implies{\sf{  \frac{x}{ - 10}  =  \frac{ -1}{10}  \:  \:  \:  \:  \:  \: , \:  \:  \frac{y}{ - 5}  =  \frac{ - 1}{10} }} \\ \\

 : \implies{\sf{  x  =  \frac{ 10}{10}  \:  \:  \:  \ \:  \:  \:   \:,  \:  \:  \:  y =  \frac{  5}{10} }} \\ \\

 : \implies{\sf{  x  =  \frac{ 1}{1}  \:  \:  \:  \ \:  \:  \:   \:  \:  \:  \:  y =  \frac{ 1}{2} }} \\ \\

 \dagger  \:  \:  \: \boxed{\sf { \purple {hence \: the \: value \: of \: x = 1 \: and \: y =  \frac{1}{2}  }}} \\ \\

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