How to solve these?

**memento mori** · August 23rd 2009, 03:38 PM

-Thanks.

**Chris L T521** · August 23rd 2009, 03:51 PM

Originally Posted by memento mori

-Thanks.

For the first one, note that when you cross multiply, you get $p(p-3)=40\implies p^2-3p=40\implies p^2-3p-40=0$ . This is now a quadratic equation which can easily be solved.

For the second one, note that when you cross multiply, you get $3(2w-3)=5w\implies 6w-9=5w$ . This is now a linear equation which can easily be solved.

For the third one, note that when you cross multiply, you get $c(c-6)=7\implies c^2-6c=7\implies c^2-6c-7=0$ . 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?

**pickslides** · August 23rd 2009, 03:51 PM

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

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

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

$p^2-3p = 40$

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

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

$p=-5,8$

**memento mori** · August 23rd 2009, 04:37 PM

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.)

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