Question

Part B
10) for what value of K, 2x + 3y =4 and (k+2) x+6y=3k +2
will have infinity many solutions?


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Answer 1

$\sf\red{\underline{\underline{Question:}}}$

For what value of K, 2x + 3y =4 and (k+2)x+6y = 3k +2 will have infinity many solutions?

$\sf\green{Given:}$

$2x+3y=4$$Here, a_1=2 ,b_1=3 ,c_1=-4$

$(k+2)x+6y=3k\\ \\Here,a_2=k+2 ,b_2=6 ,c_2=-3k+2$

$\rm\blue{For\: infinitely\: many \: solutions} \\ \\ \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2} \\ \\ \implies \frac{2}{k+2}=\frac{3}{6}=\frac{4}{3k+2}\\ \\ \implies 3(k+2)=2\times6\\ \\ \implies 3k+6=12\\ \\ \implies 6=3k\\ \\ \implies k=2$

$\sf\pink{Hence,k=2}$