-Thanks.
For the first one, note that when you cross multiply, you get $\displaystyle p(p-3)=40\implies p^2-3p=40\implies p^2-3p-40=0$. This is now a quadratic equation which can easily be solved.
For the second one, note that when you cross multiply, you get $\displaystyle 3(2w-3)=5w\implies 6w-9=5w$. This is now a linear equation which can easily be solved.
For the third one, note that when you cross multiply, you get $\displaystyle c(c-6)=7\implies c^2-6c=7\implies c^2-6c-7=0$. This is now a quadratic equation which can easily be solved.
Can you try to continue on from where I left off in each problem?
You need to post your workings when asking for help on this forum. You also need to thank any post you find helpful.
$\displaystyle \frac{p}{4} = \frac{10}{p-3}$
$\displaystyle p(p-3) = 4\times 10$
$\displaystyle p^2-3p = 40$
$\displaystyle p^2-3p - 40=0$
$\displaystyle (p+5)(p-8)=0$
$\displaystyle p=-5,8$