Question

For what values of k will the following pair of linear equations have infinitely many

solutions?

kx + 3y – (k – 3) = 0; 12x + ky – k = 0.​

Answer 1

Solution

Getting two equations with us,

→ kx + 3y - (k - 3) = 0

→ 12x + ky - k = 0

What does a infinite solution have?

Simply, all ratios are equal to each other.

→ k/12 = 3/k = - (k - 3)/(- k)

Since k/12 = 3/k

→ k² = 36

→ k = ± 6

Since 3/k = (k - 3)/k

→ 3 = k - 3

→ k = 6

Since k/12 = (k - 3)/k

→ k² = 12k - 36

→ k² - 12k + 36 = 0

→ k² - 6k - 6k + 36 = 0

→ k(k - 6) - 6(k - 6) = 0

→ (k - 6)² = 0

→ k = 6

Hence most possible value of k is + 6 and not ± 6.

Answer 2

• kx + 3y - (k - 3) = 0

• 12x + ky - k = 0

_____________ [ GIVEN ]

• We have to find the value of k if the following pair of linear equations have infinitely many solutions.

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For infinite many solutions..

» $\dfrac{ a_{1} }{ a_{2} }$ = $\dfrac{ b_{1} }{ b_{2} }$ = $\dfrac{ c_{1} }{ c_{2} }$

According to question;

Given equations are : kx + 3y - (k - 3) = 0 and 12x + ky - k = 0

Here..

$a_{1}$ = k

$a_{2}$ = 12

$b_{1}$ = 3

$b_{2}$ = k

$c_{1}$ = - (k - 3)

$c_{2}$ = - k

So,

→ $\dfrac{k}{12}$ = $\dfrac{3}{k}$ = $\dfrac{-(k\:-\:3)}{-\:k}$

→ $\dfrac{k}{12}$ = $\dfrac{3}{k}$

Cross-multiply them

→ k(k) = 12(3)

→ k² = 36

→ k = √36

→ k = ± 6 (-6 neglected)

→ k = + 6

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$\bold{\huge{k \:= \:+\: 6}}$

________ [ ANSWER ]

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