# Thread: Setting two equations equal and simplifying

1. ## Setting two equations equal and simplifying

I have 4 equations:

#1) Q1 = 8900 - P1 + 2P2
#2) Q1 = 3P1 -1200
#3) Q2 = 11,000 + 2P1 – 5P2
#4) Q2 = 14P2 - 150

I need to set equations #1 and #2 equal to each other and simplify as much as possible.

I need to do the same thing with equations #3 and #4

So,

8900 - P1 + 2P2 = 3P1 -1200

And,

11,000 + 2P1 – 5P2 = 14P2 - 150

I need to simplify each of these as much as possible. Can anyone help?? Please??

2. Originally Posted by JimmyZ11381
I have 4 equations:

#1) Q1 = 8900 - P1 + 2P2
#2) Q1 = 3P1 -1200
#3) Q2 = 11,000 + 2P1 – 5P2
#4) Q2 = 14P2 - 150

I need to set equations #1 and #2 equal to each other and simplify as much as possible.

I need to do the same thing with equations #3 and #4

So,

8900 - P1 + 2P2 = 3P1 -1200

And,

11,000 + 2P1 – 5P2 = 14P2 - 150

I need to simplify each of these as much as possible. Can anyone help?? Please??
Assuming that P1 and P2 are the same for #1-#4, we can simplify and solve this system of equations:
For simplification, I've replaced P1 = a & P2 = b.
First: #1 = #2
$\displaystyle \Rightarrow 8900 - a + 2b = 3a - 1200$
$\displaystyle \Rightarrow 8900 + 1200 = 3a +a - 2b$
$\displaystyle \Rightarrow 10100 = 4a - 2b$ we can divide by 2 to get:
$\displaystyle \Rightarrow 5050 = 2a - b$ which we will denote equation (1)

Second: #3 = #4
$\displaystyle \Rightarrow 11000 + 2a - 5b = 14b - 150$
$\displaystyle \Rightarrow 11000 + 150 = -2a + 14b + 5b$
$\displaystyle \Rightarrow 10850 = -2a + 19b$ which we will denote equation (2)

Then we attempt to solve this linear system of equations by adding the two equations:
(1) $\displaystyle 5050 = 2a - b$
(2) $\displaystyle \underline{10850 = -2a + 19b}$
$\displaystyle 15900 = 18b$
$\displaystyle \Rightarrow b = 883\frac{1}{3} = P2$

Then we can substitute this answer back into equation (1):
(1) $\displaystyle 5050 = 2a - b$
$\displaystyle \Rightarrow a = \frac{5050+b}{2} = 2966\frac{2}{3} = P1$