Question

Find the value of for which the pair of linear equations: 3(k+ 1)x + 5y − 2 = 0 & (k2 + 1)x+ (k− 2)x − 5 = 0 has no solution.

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Find the value of for which the pair of linear equations: 3(k+ 1)x + 5y − 2 = 0 & (k² + 1)x+ (k− 2)x − 5 = 0 has no solution.

ᴄᴏɴᴄᴇᴘᴛ ᴜsᴇᴅ :-

• A linear equation in two variables represents a straight line in 2D Cartesian plane .

• If we consider two linear equations in two variables, say ;

a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

Then ;

✪ Both the straight lines will coincide if ;

a1/a2 = b1/b2 = c1/c2

In this case , the system will have infinitely many solutions.

✪ Both the straight lines will be parallel if ;

a1/a2 = b1/b2 ≠ c1/c2

In this case , the system will have no solution.

✪ Both the straight lines will intersect if ;

a1/a2 ≠ b1/b2

In this case , the system will have an unique solution.

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Sᴏʟᴜᴛɪᴏɴ :-

Compare given linear pair of equations :-

⟼ (3k+1)x+3y - 2 = 0

⟼ (k²+1)x+(k - 2)y - 5 = 0

with :-

⟼ a1x + b1y + c1 = 0

⟼ a2x + b2y + c2 = 0

we get :-

⟼ a1 = (3k+1) , b1 = 3 , c1 = (-2)

⟼ a2 = (k² + 1) , b2 = (k - 2) , c2 = (-5)

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Now,

Both the straight lines will be parallel if :-

a1/a2 = b1/b2 ≠ c1/c2

In this case , the system will have no solution.

So,

➼ (3k+1)/(k² + 1) = 3/(k - 2) ≠ (-2)/(-5)

Now, comparing first 2 ,

➼ (k - 2)(3k + 1) = 3(k² + 1)

➼ 3k(k - 2)+1(k - 2) = 3k²+3

➼ 3k²- 6k + k - 2 = 3k²+3

➼ (-5k) = 3 + 2

➼ (-5k) = 5

➼ k = (-1) .

Putting This value in (2), and comparing (2) & (3) now,

➼ 3/(k - 2) ≠ (-2)/(-5)

➼ 3/((-1) - 2) ≠ (-2)/(-5)

➼ 3/(-3) ≠ (2/5)

➼ (-1) ≠ (2/5) = Satisfy.

Hence, We can Conclude That, for k = (-1) the pair of linear equations has no solution.

Answer 2

Step-by-step explanation:

$\bf \huge \: \: Question\: \: \:$

Find the value of for which the pair of linear equations: 3(k+ 1)x + 5y − 2 = 0 & (k2 + 1)x+ (k− 2)y − 5 = 0 has no solution

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Given :- Equation will have no solution.

$\bf\: \: \:(3k+1)x + 3y - 2 = 0\: \:$

$\bf\: \: \:(k^2+1)x + (k-2)y - 5 = 0\: \:$

$\bf\: \: \:3k+1/k^2+1 = 5/k-2 ≠ -2/-5\: \:$

Now,

$\bf\: \: \: 3k + 1 / k^2 + 1 = 5/k - 2\: \:$

$\bf\: \: \:⇒ ( 3k + 1 ) ( k-2 ) = 5 ( k^2+ 1 )\: \:$

$\bf\: \: \:⇒ 3k^2- 5k - 2 = 5k^2+ 5\: \:$

$\bf\: \: \:⇒ -2k^2-5k - 2 = 5\: \:$

$\bf\: \: \:⇒-2k^2-5k = 7\: \:$

$\bf⇒2k^2+5k=-7\: \:$

$\bf\: \: \:⇒2k^2+5k+7=0\: \:$

$\bf\: \: \:⇒2k^2+7k-2k+7=0\: \:$

$\bf\: \: \:⇒k(2k+7)-1(2k+7)=0\: \:$

$\bf\: \: \:⇒(k-1)(2k-7)=0\: \:$

$\bf\: \: \:⇒k =1 ,k = -7/2\: \:$

k =1 Becouse k=-7/2 is not possible