Question

If the pair of linear equations 5x+(k-4)y-20=0 and 3x+(k+7)y-12=0 has infinitely many solutions, then k is (1)Positive integer (2)Negative integer (3)Positive rational number (4)Negative rational number.

Answer 1

positive integer
it has negative which can be turned into positives

Answer 2

Answer:

The correct option is 4.

Step-by-step explanation:

The given equations are

$5x+(k-4)y-20=0$

$3x+(k+7)y-12=0$

This system of equation has many solutions if

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

$\frac{5}{3}=\frac{k-4}{k+7}=\frac{-20}{-12}$

$\frac{5}{3}=\frac{k-4}{k+7}$

$5k+35=3k-12$

$5k-3k=-35-12$

$2k=-47$

$k=-\frac{47}{2}$

It is a negative rational number. Therefore option 4 is correct.