Question

for what value of k will the pair of equation kx+3y=k-3, 12x+ky=k have no solutions

Answer 1

Heya !!!

KX + 3Y = K -3

KX + 3Y - ( K -3) = 0 ------------(1)

And,

12X + KY = K

12X + KY - K = 0

These equations are of the form of A1X + B1Y + C1 = 0 and A2X + B2Y + C2 = 0

Where,

A1 = K , B1 = 3 and C1 = -K +3

And,

A2 = 12 , B2 = K and C2 = -K

For no solution we must have,

A1/A2 = B1/B2 # C1/C2 [ Where # stand for not equal]

K / 12 = 3/K

K² = 12 × 3

K² = 36

K = ✓36 = 6

★ HOPE IT WILL HELP YOU ★

Answer 2

Answer:

k=6

Step-by-step explanation:

The given equations are:

kx+3y-(k-3)=0 and 12x+ky-k=0

Therefore, a_{1}=k, a_{2}=12, b_{1}=3, b_{2}=k, c_{1}=-(k-3), c_{2}=-k

Since, the equations has infinitely many solutions, therefore

\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}

\frac{k}{12}=\frac{3}{k}=\frac{-(k-3)}{-k}

Taking second and third equalities,

3=k-3

k=6