Question

determine the value of u so that the following pair of linear equation have no solution (3u+1)x+3y-2=0 , ( u^2 + 1 )x +(u-2)y -5=0

Answer 1

Answer:

Let Ax+By+C= (3u+1)x +3y -2 =0

Let Px+Qy+R= (u²+1)x +(u-2)y -5=0

Equation have no solution so,

A/P=B/Q

(3u+1)/(u²+1) = 3/(u-2)

(3u+1)*(u-2)=3u²+3

3u²-6u+u-2=3u²+3

-5u-2=3

-5u=5

u= -1

please mark brainliest