Question

Find the value of k for which the following pair of linear equations have infinitely many solutions. 2x+3y=7,(k-1)x+(k+2)y=3k.

Answer 1

Answer:

The value of k is 7.

Step-by-step explanation:

The given equations are

$2x+3y=7$

$(k-1)x+(k+2)y=3k$

Two equations have infinitely many solutions if

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

$\frac{2}{k-1}=\frac{3}{k+2}=\frac{7}{3k}$

$\frac{2}{k-1}=\frac{3}{k+2}$

$2(k+2)=3(k-1)$

$2k+4=3k-3$

$k=7$

$\frac{3}{k+2}=\frac{7}{3k}$

$3(3k)=7(k+2)$

$9k=7k+14$

$9k-7k=14$

$k=7$

Therefore the value of k is 7.

Answer 2

Answer:

HEY mate the above pinned image helps you to find your answer