Question

solve using cross multiplying method x+y = 7 ,2x-3y=11.

Answer 1

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Answer 2

$Let \: pair \:of \: Linear\:equations :\\a_{1}x+b_{1}y+c_{1} = 0 \:and \\a_{2}x+b_{2}y+c_{2} = 0$

$\frac{x}{(b_{1}c_{2} - b_{2}c_{1})} = \frac{y}{(c_{1}a_{2} - c_{2}a_{1})} =\frac{1}{(a_{1}b_{2} - a_{2}b_{1})}$

$Given \: pair \:of \: Linear \: Equations :$

$x + y = 7 \:implies x + y - 7 = 0 \: --(1)$

$and \: 2x - 3y = 11 \implies 2x - 3y - 11 = 0$

$Here, a_{1} = 1 , b_{1} = 1 , \: c_{1} = -7 \:and \\a_{2} = 2 , b_{2} = -3 , \: c_{2} = -11$

$\frac{x}{1(-11) -(-3)(-7)} = \frac{y}{(-7)\times 2 - (-11)\times 1 } = \frac{1}{1\times (-3) - 2 \times 1 }$

$\implies \frac{x}{-11 - 21} = \frac{y}{-14 + 11} = \frac{1}{-3- 2 }$

$\implies \frac{x}{-32} = \frac{y}{-3} = \frac{1}{-5}$

$i ) Now , \frac{x}{-32} = \frac{1}{-5}$

$\implies x = \frac{-32}{-5}$

$\implies x = \frac{32}{5}$

$ii ) \frac{y}{-3} = \frac{1}{-5}$

$\implies y = \frac{-3}{-5}$

$\implies y = \frac{3}{5}$

Therefore.,

$\green { x = \frac{32}{5} \:and \: y = \frac{3}{5}}$

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