Question

Please help with this math question​

Answer 1

For the first question, you just need to plug the values and solve for k:

$11 = \dfrac{3000}{500}+k \iff 11=6+k \iff k=11-6=5$

Now we know that $k=5$, so the cost of 200 cards will be

$y = \dfrac{3000}{200}+5 = 15+5=20$

Finally, if we want the cost to be 7 we're asking

$7=\dfrac{3000}{x}+5 \iff \dfrac{3000}{x}=2 \iff x = \dfrac{3000}{2}=1500$