Fine the value of k for which the pair of linear equation kx+3y=k+2 and 12x+ ky has no solution

Step-by-step explanation:

it will have no sol. when a1/a2=b1/b2!=c1/c2

!= means not equal to

kx+3y=k+2

kx+3y-(k+2)=0 equation 1

12x+ky=0

taking a1/a2=c1/c2

k/12=3/k

k^2=36

k=6

when k= 6 it will have no sol

22

Step-by-step explanation:

She sold two tubs of Oatmeal Raisins and it's 6 dollars a tub so we can do 6*2 or 6+6 (doesn't matter). We get $12. Then, she also sells 2 tubs of Peanut Butter, and since it's $5 a tub, then we do 5*2 or 5+5 to get 10. We add 12 and 10 (12+10) and get 22.

I'm not sure if this is right because you added $22, $24, $20, and $11 and I'm not sure what the purposes of those are.