Question

ii) 3x+y=2=0 kx+2y—3=0 and 2x-y = 3
equations are consistent​

Answer 1

Answer:

Step-by-step explanation:

The given lines are:

3x+y−2=0.(i)

kx+2y−3=0(ii)

2x−y−3=0(iii)

On solving (i) and (iii) by cross multiplication, we get

x(−3−2)=y(−4+9)=1(−3−2)⇒x−5=y5=1−5

⇒x=(−5−5)=1andy=(5−5)=−1

Thus, the point of intersection of (i) and (iii) is P(1,-1)

For the given lines to intersect at a point ,x=1 and y=-1 must satisfy (ii) also.

∴(k×1)+2×(−1)−3=0⇒k−5=0⇒k=5

Hence, k=5

Answer 2

Step-by-step explanation:

Let , 3x+y=2 equation 1sr

and 2x-y =3 equation 2nd

Adding both equations we get,

x=1 and y= -1

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

k-2=3

k=3+2

k=5