Question

For what value of k does the system equation x+2y=3 and 5x+ky+7=0 has no salutation.​

Answer 1

Answer:

k=10

Step-by-step explanation:

here a1=1,b1=2,c1=-3

a2=5,b2=k,c2=7

for no solution-

a1/a2=b1/b2≠c1/c2

we take a1/a2=b1/b2

1/5=2/k

k=10

Answer 2

Step-by-step explanation:

$x + 2y = 3 \\ 2y = 3 - x \\ y = \frac{3 - x}{2 } \\ 5x + ky + 7 = 0 \\ 5x + k( \frac{3 - x}{2} ) = 0 \\ 5x + \frac{3k - kx}{2} = 0 \\ 10x + 3k - kx = 0 \\ 10x + 3k = kx \\ \frac{10x + 3k}{x } = k \\ \frac{10x}{x} + \frac{3k}{x} = k \\ \frac{10}{1} + \frac{3k}{x} = k \\ 10x + 3k = k \\ 10x + 3k - k = 0 \\ 10x - 2k = 0 \\ 5x - k = 0 \\ 5x = k \\ k = 5x$