9a+6b=3

We must start by cancelling out one of the variables by manipulating them so they both have the same polynomial in them.

 

9a + 6b = 3      [/3

3a + 2b = 1     [*5

15a + 10b = 5

 

25a + 35b = 5    [/5

5a + 7b = 1    [*3

15a + 21b = 3

 

Now they both have 15a in them, we can take one from the other

 

   15a + 21b = 3

-  15a + 10b = 5

----------------------

            11b = -2

              b = -2/11

 

Now find what a is by substituting in the value of b:

 

9a + 6b = 3

9a = 3 - 6b

a = (1 - 2b) / 3         , b = -2/11

a = (1 - 2(-2/11))/3

a = (1+4/11)/3 = (15/11)/3 = 5/11

a = 5/11

 

To check our anwsers, substitute in the values for a and b

9a + 6b = 3

9*5/11 + 6*-2/11 = 3, good

25a + 35b = 5

25*5/11 + 35*-2/11 = 5, good

all correct. so the solutions are:

a = 5/11

b = -2/11

 

From Sam of Herefordshire, England