Question

7. Find the value of m, when (m+1)x=3ky+15=0 and 5x+ky+5=0 are coincident..​

Answer 1

Given equations :-

(m + 1)x = 3ky + 15 = 0

5x + ky + 5 = 0

For a pair of linear equation to be coincident,

a1/a2 = b1/b2 = c1/c2

Here,

a1/a2 = (m + 1)/5

b1/b2 = 3k/k

c1/c2 = 15/5

On comparing,

(m + 1)/5 = 3k/k = 15/5

Finding m,

(m + 1)/5 = 15/5

5(m + 1) = 15(5)

5m + 5 = 75

5m = 75 - 5

5m = 70

m = 70/5

m = 14

Answer 2

◇ qUESTIOn

$\tt{Find \:the\: value\: of\: <strong>m</strong>,\: when (m+1)x=3ky+15=0\\\tt and 5x+ky+5=0\: are \:coincident.}$

◇sOLUTIOn

$\rightarrow\tt(m+1)x=3ky+15=0\\ \rightarrow\tt 5x+ky+5=0\\ \rightarrow \tt a1=(m+1), b1= 3k, c1=15\\\rightarrow\tt a=5 , b2= k , c2= 5$

$\implies \tt\frac{a1}{a2}=\frac{b1}{b2}=\frac{c1}{c2}\\ \implies \tt\frac{m+1}{5}=\frac{3k}{k}=\frac{15}{5}$

Now finding the value of 【 "m"】

$\implies \tt\frac{m+1}{5}=\frac{15}{5}\\ \implies\tt 5(m+1)=15×5\\ \implies\tt 5m+5=75 \\ \implies \tt m=\frac{70}{5}\\\implies\tt m=14 \\ \implies\tt \red{\underline{\fbox{m=14} }}$