Question

Find the value(s) of k for which the given pair of linear equations has unique

solution: 3x + y = 5 ; kx + 4y = 6​

Answer 1

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GIVEN THAT:-

$&#10144$ The pair of linear equations

3x + y = 5 and kx + 4y = 6 has unique solution.

SOLUTIONS:-

$&#10144$ Comparing the both equaton

3x + y = 5 and kx + 4y = 6 with standard equaton

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

we get

a1 = 3, b1 = 1 , c1 = -5

a2 = k , b2 = 4 , c2 = -6

$&#10144$ Since both equaton has unique solution, so

$&#10230 \: \: \frac{ a_{1} }{a_{2}} \cancel= \frac{b_{1}}{b_{2}} \\ \\ &#10230 \: \: \frac{3}{k} \cancel= \frac{1}{4} \: \: \: \\ \\ &#10230 \: \: k \: \cancel=12 \: \:$

• The value of k are all real numbers expect 12.