Question

for what value of k, 2x+3y=4 and (k+2)x+6y=3k+2 will have infinitely many solutions

Answer 1

We know for having infinitely many solutions for two linear equations.namely
ax+by=c
dx+ey=f
Then the condition is
a
$\frac{a}{d} = \frac{c}{e}$
Therefore
From the given question
a=2.b=3.c=(k+2).d=6.
$\frac{2}{k + 2} = \frac{3}{6}$
$2 \times \frac{6}{3} = k + 2$
$4 = k + 2$
Therefore k=2.