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Answer 1

In the last part, ....


K= +6 or -6

if k= 6, then k is not equal to √(18-3×6
√0= 0


so , k= 6 satisfies both equations.

The first 2 parts have been solved in the image.

Hope it helps you

Answer 2

$\large{\color{orange}{\bf HELLO \: \: DEAR,}}$

$\bf \: condition \: \: for \: \: no \: \: solution :- \frac{a_1}{b_1} = \frac{b_1}{b_2} \neq\frac{c_1}{c_2}$

$\boxed{\bf 1} . \bf \: x + 2y = 0\\ \\ \: \: \: \: \: \bf 2x + ky = 5$

$\bf \: a_1 = 1 , b_1 = 2 , a_2 = 2 , b_2 = k$

$\bf \: \frac{1}{2} = \frac{2}{k} \\ \\ \bf \: k = 2*2 \\ \\ \bf \: k = 4$

$\boxed{\bf 2}. \bf2x - ky + 3 = 0\\ \\ \bf3x + 2y - 1 = 0 \\ \\ \bf \: a_1 = 2 , b_1 = 3 , a_2 = (-k) , b_2 = 2 a_1/a_2 = b_1/b_2\\ \\ \bf \: 2/-k = 3/2 \\ \\ \bf \:-k = 4/3$

$\boxed{\bf 3}. \:\: \bf kx + 3y = k - 3 \\ \\ \: \: \: \: \bf 12x + ky = 6$

$\bf \: a_1 = k , b_1 = 3 , a_2 = 12 , b_2 = k \\ \\ \bf \: a_1/a_2 = b_1/b_2 k/12 = 3/k\\ \\ \bf k^{2} = 36 \\ \\ \bf k = +_-6$

$\color{blue}\underline{\bf I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}$