Any non-integer solutions should be given as fractions in their simplest form.

5a-b=9

a+6b=-3

Also, please show working.

- Oct 23rd 2010, 05:36 AMmathsstudentPlease help me solve the following pair of simultaneous equations algebraically.
Any non-integer solutions should be given as fractions in their simplest form.

5a-b=9

a+6b=-3

Also, please show working. - Oct 23rd 2010, 05:39 AMProve It
Start by multiplying the first equation by 6. Then add the two equations together to eliminate $\displaystyle b$.

- Oct 23rd 2010, 05:49 AMmathsstudent
hi,

what's next? - Oct 23rd 2010, 05:52 AMProve It
Why not show what you have done and where you are stuck?

- Oct 23rd 2010, 06:01 AMmathsstudent
I multiplied the first equation by 6

and added the 2 equations together

30a - b = 54

a+6b = -3

31 + b =51 - Oct 23rd 2010, 06:02 AMProve It
It should actually have been

$\displaystyle 30a - 6b = 54$

$\displaystyle a + 6b = -3$.

Now try adding the equations together. What do you get? - Oct 23rd 2010, 06:06 AMmathsstudent
Is it

31a+12b=51?

Thanks - Oct 23rd 2010, 06:10 AMUnknown008
No, adding them gives:

(30a - 6b) + (a + 6b) = 54 -3

31a = 51

b - b = 0