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# Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method. x - 3y - 7 = 0 3x - 3y - 15 = 0

Q1.    Which of the following pairs of linear equations has unique solution, no solution, or infinitely many solutions. In case there is a unique solution, find it by using cross multiplication method.

(iv)    $\\x - 3y -7 = 0\\ 3x -3y -15 =0$

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Given the equations,

$\\x - 3y -7 = 0.........(1)\\ 3x -3y -15 =0........(2)$

Comparing these equations with $a_1x+b_1y+c_1=0\:and\:a_2x+b_2y+c_2=0$, we get

$\frac{a_1}{a_2}=\frac{1}{3}$

$\frac{b_1}{b_2}=\frac{-3}{-3}=1$

$\frac{c_1}{c_2}=\frac{-7}{-15}=\frac{7}{15}$

As we can see,

$\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\neq\frac{c_1}{c_2}$

Hence, the pair of equations has exactly one solution.

By Cross multiplication method,

$\frac{x}{b_1c_2-b_2c_1}=\frac{y}{c_1a_2-c_2a_1}=\frac{1}{a_1b_2-a_2b_1}$

$\frac{x}{(-3)(-15)-(-3)(-7)}=\frac{y}{(-7)(3)-(-15)(1)}=\frac{1}{(1)(-3)-(3)(-3)}$

$\frac{x}{45-21}=\frac{y}{-21+15}=\frac{1}{-3+9}$

$\frac{x}{24}=\frac{y}{-6}=\frac{1}{6}$

$x=\frac{24}{6}=4,\:and\:y=-1$

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