Question 133632
In order to perform the addition method, the coefficient on one of the variables in one equation must be the additive inverse of the coefficient on the same variable in the other equation.


The lowest common multiple of 2 and 3 (the coefficients on a) is 6, so we need to multiply the first equation by 3 so that the coefficient on a is 6 and multiply the second equation by -2 so that the coefficient on a is the additive inverse of 6, or -6.  Or vice-versa (1st equation by -3, second by 2).


{{{2a + 3b = -1}}}
{{{3a + 5b = -2}}}


{{{6a + 9b = -3}}}
{{{-6a -10b = 4}}}


Now you can add the two left sides together and set the result equal to the sum of the right sides.  That is because if {{{a=b}}} and {{{c=d}}}, then {{{a+c=b+d}}}


{{{6a-6a+9b-10b=-3+4}}}
{{{-b=1}}}
{{{b=-1}}}


Now you can do either of two things.  You could take this newly derived value for b and substitute it into either of the original equations and then solve the result for a, or you could recognize that the LCM for 5 and 3 is 15, then multiply eq 1 by 5 and eq 2 by -3, add the equations and determine a that way.  Since b came out to be a nice tidy integer, let's do the substitution thing.


{{{2a + 3(-1) = -1}}}
{{{2a=-1+3}}}
{{{2a=2}}}
{{{a=1}}}


Therefore the solution set is the ordered pair (a,b)=(1,-1)


Check your answer:
{{{2(1) + 3(-1) =2-3= -1}}}
{{{3(1) + 5(-1) =3-5= -2}}}  
The derived solution makes both equations true, answer checks.