Question

please solve this question
find the value of x and y by elimination method ​

Answer 1

Solution :

$\sf 2x - 3y - 3 = 0 \: . \: . \: . \: . \: . \: . \: .(1)$

$\sf \frac{2x}{3} + 4y + \frac{1}{2} = 0\\ \\ \sf \frac{4x + 24y + 3}{6} = 0 \\ \\ \sf4x + 24y + 3 = 0 \: . \: . \: . \: . \: . \: .(2)$

Multiply ( 1 ) by 2

$\sf 2(2x - 3y - 3 = 0) \\ \\ \sf 4x - 6y - 6 = 0 \: . \: \: . \: . \: . \: . \: .(3)$

Now by elimination method

$\: \: \: 4x + 24 + 3 = 0 \\ \: \: \: 4x - 6y - 6 = 0 \\ \underline{ - \: \: \: \: \: + \: \: \: \: \: \: + \: \: \: \: \: \: \: \: \: \: \:\: } \\ \: \: \: \: \: \: \: \: \: \: \: \: \sf30y + 9= 0 \\ \\ \implies\sf 30y = - 9 \\ \\ \implies\sf y = - \frac{9}{30} \\ \\\large \implies \boxed{\boxed{ \sf y = - \frac{3}{10}}}$

Put value of y in ( 1 )

$\sf \implies2x - 3y - 3 = 0 \\ \\\sf \implies 2x - 3 \times \big(- \frac{3}{10} \big) - 3 = 0 \\ \\\sf \implies 2x + \frac{9}{10} - 3 = 0 \\ \\\sf \implies 2x + \frac{9 - 30}{10} = 0 \\ \\ \sf \implies 2x = \frac{21}{10} \\ \\ \large\implies \boxed{ \boxed{ \sf x = \frac{21}{20} } }$

Answer 2

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