# Rearranging eqn

1. Mar 1, 2008

### 2slowtogofast

a1 + a3 = ( b2*(b1+b3) ) / (b1 + b2 + b3)

a1 + a2 = ( b3*(b1+b2) ) / (b1 + b2 + b3)

a2 + a3 = ( b1*(b2+b3) ) / (b1 + b2 + b3)

those equations are given

i need to show that

a1 = (b2 * b3) / (b1 + b2 + b3)

i just need some help starting this off i think i need to solve for a1 and then substitute but what ever i do i am always left with a2 or an a3 in my answer any help is appreciated

2. Mar 1, 2008

### arildno

Subtract the third from the second, and add the resulting eq to the first, and multiply the result with 1/2

Last edited: Mar 1, 2008
3. Mar 1, 2008

### cristo

Staff Emeritus
Try subtracting the third equation from the second. This will leave you with two equations with left hand sides (a1+a3) and (a1-a3), which you may see how to solve for a1.

4. Mar 1, 2008

### 2slowtogofast

thanks for the advice im going to try it now

5. Mar 1, 2008

### 2slowtogofast

i am just unclear on why you multiply by 1/2

6. Mar 1, 2008

### arildno

Because prior to thar move, you'll have an equation with 2a1 on your left-hand side.
Multiplying the equation with 1/2 eliminates that factor of 2 from the left hand side, leaving a1 on that side.

7. Mar 1, 2008

### 2slowtogofast

ok now i got 2a1 = 2b2*b3 / b1 + b2 + b3

multiply by 1/2 and i get the answer i understand how i did this. but what i am asking is every time i do a problem like this do i multiply by 1/2 at the end or it just worked out that way this time? thanks for the help i havent done these in two years so i am a bit rusty

8. Mar 1, 2008

### Tedjn

It depends on how many equations there are and what those equations are. Just make sure when you add equations, you add both sides. What you will then get is a new equation that is true; if it ends up that you have 2a1, then divide both sides by 2 (multiply by 1/2), if you have 3a1, then divide both sides by 3, etc.

9. Mar 1, 2008

### 2slowtogofast

i just want to get more practice with this if i wanted to show :

b1 = (a1*a2 + a2*a3 + a3*a1 ) / a1

would i solve all three eqns for

b1 + b3 eqn 1

b1 +b2 eqn 2

b2 + b3 eqn 3

instead of having a1 + a3 and so forth and then you the same approach i used for my first problem?

10. Mar 1, 2008

### Tedjn

If that's how the equations look, sure. It doesn't matter what the letter is--it's just a place holder and could be called a, b, c, d, *, \$ or whatever.