# Thread: Solving this system of equation with two unknowns.

1. ## Solving this system of equation with two unknowns.

Hello,

I am trying to solve:

a0 = 2 = a · 30 + b · 20 = a + b
a1 = 6 = a · 31 + b · 21 = 3a + 2b

2. ## Re: Solving this system of equation with two unknowns.

Hey Archytas.

Hint - Try getting one equation in terms of one variable and substituting it into the other one.

3. ## Re: Solving this system of equation with two unknowns.

Or- multiply the second equation by 2 and subtract that from the first equation.

Then multiply the second equation by 3 and subtract the first equation from that.

4. ## Re: Solving this system of equation with two unknowns.

Originally Posted by Archytas
a0 = 2 = a · 30 + b · 20 = a + b
a1 = 6 = a · 31 + b · 21 = 3a + 2b
Looks to me that above is simply this:
a + b = 2
3a + 2b = 6
...which evidently means b = 0.

Do I need another coffee?

5. ## Re: Solving this system of equation with two unknowns.

I am no so sure that is it "evident". yes, if a= 2 then 3a= 6 so a= 2, b= 0 satisfy the equation.

Or, as chiro suggested, from a+ b= 2, b= 2- a so the second equation becomes 3a+ 2(2- a)= a+ 4= 6 so a= 2 and then b= 2- 2= 0.

Or, as I suggested, multiplying the first equation by 2, 2a+ 2b= 4 and, subtracting that from 3a+ 2b= 6, a= 2. Multiplying the first equation by 3, 3a+ 3b= 6 and, subtracting that from 3a+ 2b= 6 from that b= 0.

6. ## Re: Solving this system of equation with two unknowns.

Originally Posted by HallsofIvy
I am no so sure that is it "evident".
.........
Multiplying the first equation by 3, 3a+ 3b= 6 and,
subtracting that from 3a+ 2b= 6 from that b= 0.
That's why it's evident (more or less).

7. ## Re: Solving this system of equation with two unknowns.

It is evident thanks to your clarification, thank you!