Question

find the value of k for which the following pair of linear equation has a unique solution 2x+3y=7;(k-1)x+(k+2)y=3k

Answer 1

Answer:

$For\:all\:real\:values\: of \:k ,\\except \:k≠7\:given\\equations\:have \:unique\: solution$

Step-by-step explanation:

$Given ,\: pair \:of\:linear\:equations :\\2x+3y-7=0---(1)\\(k-1)x+(k+2)y-3k=0---(2)$

$Compare\: these \: equations\\with\:a_{1}x+b_{1}y+c_{1}=0\\and,a_{2}x+b_{2}y+c_{2}=0$

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

$\frac{a_{1}}{a_{2}}≠\frac{a_{1}}{a_{2}}\\(Given\: equations\:have\\unique\: solution)$

$\implies \frac{2}{k-1}≠\frac{3}{k+2}$

$\implies 2(k+2)≠3(k-1)$

$\implies 2k+4≠3k-3$

$\implies 4+3≠3k-2k$

$\implies 7≠k$

Therefore,

$For\:all\:real\:values\: of \:k ,\\except \:k≠7\:given\\equations\:have \:unique\: solution$

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Answer 2

Answer:

The value of k is 7.

Step-by-step explanation:

Let the equation (i) be 2x + 3y – 7 = 0  

Let the equation (ii) be (k – 1)x + (k + 2)y – 3k = 0  

$\begin{array}{l}{a_{1}=2, b_{1}=3, c_{1}=-7} \\ {a_{2}=(k-1), b_{2}=(k+2), c_{2}=-3 k}\end{array}$

We have,

$a_{1} a_{2}=b_{1} b_{2}=c_{1} c_{2}$

For infinite number of solutions

2k–1=3k+2=−7−3k

2k–1=3k+2=73k

Now, the following cases arises

Case 1:

2k–1=3k+2

2(k+2)=3(k–1)⇒2k+4=3k−3

k=7

Case 2:

3k+2=73k

(k+2)=9k⇒7k+14=9k

k=7

Case 3:

2k–1=73k

7k–7=6k

k=7