Question

For what value of k, does the pair of linear equations kx – y = 2, 192x – 64y = 128 has infinite

solution ?

THOSE WHO KNOW THE ANS GIVE ​

Answer 1

SOLUTION

TO DETERMINE

The value of k for which the pair of linear equations

kx – y = 2, 192x – 64y = 128 has infinite solution

CONCEPT TO BE IMPLEMENTED

A pair of linear equations

$\sf{a_1x + b_1y = c_1 \: \: and \: \:a_2x + b_2y = c_2 }$

has infinite number of solutions if

$\displaystyle \sf{ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} }$

EVALUATION

Here the given pair of linear equations are

$\sf{kx - y = 2 \: \: \: .........(1)}$

$\sf{192x - 64y = 128 \: \: ......(2)}$

Equation (2) can be rewritten as

$\sf{3x - y = 2 \: \: ......(3)}$

Comparing equation (1) & (3) with

$\sf{a_1x + b_1y = c_1 \: \: and \: \:a_2x + b_2y = c_2 } \: \: we \: get$

$\sf{a_1 = k, b_1 = - 1 ,c_1 = 2 \: and \: a_1 = 3, b_2 = - 1, c_2 = 2}$

Now for infinite number Solutions we have

$\displaystyle \sf{ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} }$

$\displaystyle \sf{ \implies \frac{k}{3} = \frac{ - 1}{ - 1} = \frac{2}{2} }$

$\displaystyle \sf{ \implies \frac{k}{3} = 1 }$

$\displaystyle \sf{ \implies k = 3 }$

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