1. I need help to find the values of "p" & "c" if:

pc=120
(p+2)(c-3)=120

My first step was to expand the brackets to get:

pc=120
pc-6-3p+2c=120

Which simplifies to:

pc=120 A
pc-3p+2c=126 B

I then take A from B which leaves:

-3p+2c=6

I am not sure what to do from this point nor am i confident it is correct up to this point. I would be very greatful for any pointers.

2. Simultaneous equations

Hello tomh381

Your working is perfect up to this point. Thanks for showing us what you've done so far.

All you need to do now is to make $c$ the subject of your last equation:

$c = \tfrac{1}{2}(6 + 3p)$ (1)

A to eliminate $c$. You'll then get a quadratic equation in $p$, which will factorise to give you two values. Substitute back into equation (1) here, and you're done.

Can you complete this now?

3. I am not sure I can, this is what i have so far:

cp=120 A
c=(6+3p)/2 B

I then minus B from A to leave:

p=120-((6+3p)/2)

I am unsure on what to do from here!

Thanks_381

4. cp=120 --------- Eqn 1
c=(6+3p)/2 ----- Eqn 2

Sub Eqn 2 into Eqn 1

(6+3p)/2 )p=120
(6p+3p^2)/2=120
6p+3p^2 = 240
3p^2+6p-240=0

Can you take it from here and solve for p. Hope it helps.

Originally Posted by tomh381
I am not sure I can, this is what i have so far:

cp=120 A
c=(6+3p)/2 B

I then minus B from A to leave:

p=120-((6+3p)/2)

I am unsure on what to do from here!

Thanks_381