Question 124242
{{{x=3y+11}}}
{{{2x+5y=0}}}


Substitution:
The first equation is already solved for x, so you can take the right-hand expression and substitute it for x into the second equation:


{{{2x+5y=0}}}
{{{2(3y+11)+5y=0}}}


Simplify and solve:
{{{6y+22+5y=0}}}
{{{11y+22=0}}}
{{{11y=-22}}}
{{{y=-2}}}


Now that you have a value for y, you can substitute that back into either equation to solve for x:
{{{x=3(-2)+11}}}
{{{x=-6+11}}}
{{{x=5}}}


So the solution set for this system is the ordered pair (5,-2)


Elimination:
Your second equation is in standard form ({{{ax+by=c}}}), so put the first equation into standard form as well.

{{{x=3y+11}}}
{{{x-3y=11}}}


1. {{{x-3y=11}}}
2. {{{2x+5y=0}}}


The idea is to multiply one or both of the equations by a constant or constants so that the coefficients on one of the variables will become additive inverses.  In this case, multiplying the first equation by -2 will give you a -2x in the first equation and 2x in the second equation.


3. {{{-2x+6y=-22}}}
2. {{{2x+5y=0}}}


Now add the two equations, term by term:


4. {{{0x+11y=-22}}}


And solve:


{{{11y=-22}}}
{{{y=-2}}}


Now you can do either of two things.  One, you can substitute this value for y back into either of the original equations and solve for x.  Or two, you can go back to the original equations, find a different multiplier or multipliers that will allow you to eliminate the y variable and solve for x.


We've already done the first way as the second step of the substitution method, so let's try the second way:


1. {{{x-3y=11}}}
2. {{{2x+5y=0}}}


Multiply the first equation by 5 and the second equation by 3


3. {{{5x-15y=55}}}
4. {{{6x+15y=0}}}


Add the equations, term-by-term:


5. {{{11x+0y=55}}}


Solve
{{{11x=55}}}
{{{x=5}}}


And again, the solution set is the ordered pair (5,-2).  I sincerely hope that you weren't surprised that we got the same answer with either method.