1. How to solve these?

-Thanks.

2. Originally Posted by memento mori

-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?

3. You need to post your workings when asking for help on this forum. You also need to thank any post you find helpful.

$\displaystyle \frac{p}{4} = \frac{10}{p-3}$

$\displaystyle p(p-3) = 4\times 10$

$\displaystyle p^2-3p = 40$

$\displaystyle p^2-3p - 40=0$

$\displaystyle (p+5)(p-8)=0$

$\displaystyle p=-5,8$

4. Thank you both very much; it's now much easier for me to solve these sorts of problems. (I'll also keep that in mind, pickslides.)