Question

Solve the pair of linear equations :
7x-4y=49 5x-6y=57 by cross multiplication method

Answer 1

the answer is x=3 and y=-7

Answer 2

$If \: a_{1}x+b_{1}y +c_{1} = 0 ,\\and \: a_{2}x+b_{2}y +c_{2} = 0 \:are \: pair \:of \:linear \\equations$

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

$Given \: Linear \:pair :$

$7x - 4y = 49 \implies 7x - 4y - 49 = 0\: --(1)$

$and \: 5x - 6y = 57 \implies 5x - 6y - 57 \: --(2)$

$Here , a_{1} = 7, \: b_{1} = -4 \:and \: c_{1} = -49$

$and , a_{2} = 5, \: b_{2} = -6\:and \: c_{2} = -57$

$\frac{x}{(-4)(-57)-(-6)(-49)} = \frac{y}{(-49)\times 5 - (-57)\times 7} = \frac{1}{7(-6) - 5 \times (-4) }$

$\implies \frac{x}{228-294} = \frac{y}{-245+399} = \frac{1}{-42+20}$

$\implies \frac{x}{-66} = \frac{y}{154} = \frac{1}{-22}$

$i )Now, \frac{x}{-66} = \frac{1}{-22}$

$\implies x = \frac{-66}{-22}$

$\implies x = 3$

$ii) \frac{y}{154} = \frac{1}{-22}$

$\implies 5 = \frac{154}{-22}$

$\implies y = -7$

Therefore ,

$\green { x = 3 \: and \: y = -7 }$

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