How to Solve for a1 in Given Equations with Multiple Variables and Fractions

[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

 

[arildno]

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

 

[cristo]

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.

 

[2slowtogofast]

thanks for the advice I am going to try it now

 

[2slowtogofast]

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

 

[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.

 

[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 haven't done these in two years so i am a bit rusty

 

[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.

 

[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?

 

[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.