solve system of equations

Solve the following system:
{y = (5 x)/6+3 |     (equation 1)
y = x/9+1 |     (equation 2)
Express the system in standard form:
{-(5 x)/6+y = 3 |     (equation 1)
-x/9+y = 1 |     (equation 2)
Subtract 2/15 × (equation 1) from equation 2:
{-(5 x)/6+y = 3 |     (equation 1)
0 x+(13 y)/15 = 3/5 |     (equation 2)
Multiply equation 1 by 6:
{-(5 x)+6 y = 18 |     (equation 1)
0 x+(13 y)/15 = 3/5 |     (equation 2)
Multiply equation 2 by 15:
{-(5 x)+6 y = 18 |     (equation 1)
0 x+13 y = 9 |     (equation 2)
Divide equation 2 by 13:
{-(5 x)+6 y = 18 |     (equation 1)
0 x+y = 9/13 |     (equation 2)
Subtract 6 × (equation 2) from equation 1:
{-(5 x)+0 y = 180/13 |     (equation 1)
0 x+y = 9/13 |     (equation 2)
Divide equation 1 by -5:
{x+0 y = (-36)/13 |     (equation 1)
0 x+y = 9/13 |     (equation 2)
Collect results:
Answer: |  x = -36/13       y = 9/13

y=5/6x+3 y=1/9x+1

Thanks guest, my answer is the same as yours, I just thought it might be easier to comprehend.

Why don't you try to use a little LaTax coding.  It is not that hard to get started and then you can build from there.

For instance,

The coding for a 1/2  is   \frac{1}{2}

Multiply the first equation by 6 and the second one by 9 to get rid of the fractions.

\(6y=5x+18 \qquad(1a)\\ 9y=x+9 \qquad (2a) \rightarrow \qquad x=9y-9 \qquad(2b)\\ \mbox{Sub (2b) into (1a)}\\ 6y=5(9y-9)+18\\ 6y=45y-45+18\\ 6y=45y-27\\ 27 =45y-6y\\ 27 =39y\\ 9 =13y\\ y=\frac{9}{13}\\ x=9(\frac{9}{13})-9\\ x=\frac{81}{13}-\frac{117}{13}\\ x=\frac{-36}{13}\\ x=\;{-2\frac{10}{13}}\)