Question

The value of k for which the pair of linear equations 3x+4y+5=0 and 12x+2ky+17=0 has no solution

Answer 1

Step-by-step explanation:

hope the above answer will help you

Answer 2

Given :-

$\qquad$☀Linear pair equations :

• 3x + 4y + 5 = 0
• 12x + 2ky + 17 = 0

To Find :-

• Value of K

Solution :-

$\qquad$☀We have two linear pair equations :-

• 3x + 4y + 5 = 0
• 12x + 2ky + 17 = 0

Where :-

$\qquad\quad \pmb {\mathfrak{a_1 = 3}}$

$\qquad\quad \pmb {\mathfrak{b_1 = 4 }}$

$\qquad\quad \pmb {\mathfrak{b_2 = 2k}}$

$\qquad\quad \pmb {\mathfrak{c_1 = -5}}$

$\qquad\quad \pmb {\mathfrak{c_2 = -17}}$

Substituting the values :-

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{a_1}{a_2} =\dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2} }}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{3}{12} =\dfrac{4}{2k} \neq \dfrac{-5}{-17} }}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{3}{12} =\dfrac{4}{2k} }}$

$\qquad\leadsto\quad \pmb {\mathfrak{\dfrac{3}{12} =\dfrac{4}{2k}}}$

$\qquad\leadsto\quad \pmb {\mathfrak{6k = 48}}$

$\qquad\leadsto\quad \pmb {\mathfrak{k=\cancel{\dfrac{48}{6}}}}$

$\pink{\qquad\leadsto\quad \pmb {\mathfrak{k=8}}}$

$\therefore\:\underline{\textsf{Value of k is \textbf{ 8 }}}.$