Question

How to combine two linear equations together?

Answer 1

How is this actually done? Can someone explain this?

Answer 2

Q_0=Q_1+JA

Q_0c_0=Q_1c_1

solving the second equation for Q1
Q_0c_0/c_1=Q_1

substituting this into the first
Q_0=Q_0c_0/c_1+JA

rearranging
Q_0-Q_0c_0/c_1=JA
Q_0(1-c_0/c_1)/A=J

Answer 3

Would it not be (Q_0*c_1)/(Q_0*c_0)=JA ??

Answer 4

how'd you get that?

Answer 5

If Q1=(Q0*c0)/c1

Sub that back into the original equation.

Q0=((Q0*c0)/c1)+JA

Answer 6

Multiple and divide across to move the fraction to the LHS to get JA on the RHS

Answer 7

Then divide the entire equation by A to get the equation in terms of J

Answer 8

Q0=((Q0*c0)/c1)+JA

when multiplying or dividing both sides of the equation , you have to apply the operation to all term on both side

if you multiply both sides by c1, and divide by Q0C0

you would get

Q0c1=Q0*c0+JAc1

Q0c1/Q0*c0=1+JAc1/Q0*c0

which doesn't really help,

Answer 9

you want to get the Q0's together on one side , by misusing, before you multiply or divide

Answer 10

So basically what you're saying where I was going wrong is that when I am this step as;

Q0=((Q0*c0)/c1)+JA


I just substract ((Q0*c0)/c1) from both sides of the equation?

Answer 11

yeah this line is right Q0=(Q0*c0/c1)+JA

now takeaway (Q0*c0/c1) form both sides

Answer 12

*from

Answer 13

I'll get J=Q0-((Q1-c0)/((c1*A))

Answer 14

No wait

Answer 15

I'll get J=Q0-((Q1*c0)/((c1*A))

Answer 16

Q0=(Q0*c0/c1)+JA

Q0-(Q0*c0/c1)=JA

J=[Q0-(Q0*c0/c1)]/A

Answer 17

Oh yeah the entire equation by A

Answer 18

now both the terms in the square brackets have a common factor of Q0, so you can factor it out

J=Q0[1-(c0/c1)]/A

Answer 19

I see, it's my simplifying that's the problem..

Answer 20

now
Q0[1-(c0/c1)]/A is the same as (Q0/A)[1-(c0/c1)]

and that's the target.

Answer 21

Thank you for the help. Annoying little question.

Answer 22

Remember that when applying an operation to an equation, you have to apply it to all the terms on both sides to keep the equation balanced