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