Question

2x+3y=7(k-1) x+(k+2)y=3k find the value of k

Answer 1

if a1x+b1y=a2x+b2y then a1/a2=b1/b2
2/(k-1)=3/(k+2)
2(k+2)=3(k-1)
2k+4=3k-3
4+3=3k-2k
7=k

Answer 2

Answer:

$Value \:of \:k = \frac{-1}{2}$

Step-by-step explanation:

Given system of equations:

2x+3y-7(k-1)=0 ----(1)

x+(k+2)y-3k=0 ----(2)

$Compare \: above \: equations \\with \:a_{1}x+b_{1}y+c_{1}=0\:and\\a_{2}x+b_{2}y+c_{2}=0\:we\:get$

$a_{1}=2\:b_{1}=3\:c_{1}=-7(k-1);\\a_{2}=1\:b_{2}=(k+2)\:c_{2}=3k;$

/* If two lines are parallel,then

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}$

$\implies \frac{2}{1}=\frac{3}{k+2}$

$\implies 2(k+2)=3$

$\implies 2k+4=3$

$\implies 2k = 3-4$

$\implies 2k = -1$

$\implies k = \frac{-1}{2}$

Therefore,

$Value \:of \:k = \frac{-1}{2}$

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