Question 131817
{{{3b + 2c = 46}}}
{{{5c + b = 11}}}


Things are out of order here because the 'b' terms aren't lined up and neither are the 'c' terms.  If you put either of the equations in slope-intercept form, you are just going to make a bigger mess of things.  What you need to do first is to use the commutative property of addition to swap the position of the terms on the left side of either of the equations so that 'b' in the second equation is lined up under '3b' in the first, etc.


{{{3b + 2c = 46}}}
{{{b + 5c = 11}}}


Now you are asked to solve the system by the method of linear combination, aka 'the addition method'  I like to call this method the 'elimination' method because we are going to perform some operations that will eliminate one of the variables.

What we need is a number that can be multiplied across one of the equations so that the coefficient on one of the variables in that equation becomes the additive inverse of the coefficient on that same variable in the other equation.  (remember that the additive inverse is the number you can add to any number and get zero as a result.  -2 and 2 are additive inverses of each other.)

In this case, we can multiply the second equation by -3.  Let's do that.
{{{-3b-15c=-33}}}.  Then we line up the first equation underneath it.


{{{-3b-15c=-33}}}
{{{3b + 2c = 46}}}

And then add the two equations together, term by term:
{{{0b-13c=13}}}
{{{-13c=13}}}
{{{c=-1}}}


There are two ways to go from here.  We can either take this value for c that we just discovered and substitute it back into either of the original equations allowing us to solve for the other variable, b, or we can go back to the original equations and find a way to eliminate the c variable, thus solving for b the same way we just solved for c.  I'm going to demonstrate this second method because there is a little fly in the ointment that you need to see to learn to solve these types of problems in general.


{{{3b + 2c = 46}}}
{{{b + 5c = 11}}}


We could multiply the second equation by {{{-2/5}}} to make the coefficient on c be -2, but that would make for some very messy numbers on the other coefficients that could cause difficulty when we go to add the two equations.  A better method is to find a common multiple between the coefficients on c in the two equations and multiply by appropriate integers to make the c coefficients be additive inverses.  Here, both 2 and 5 are factors of 10, so if we multiply the first equation by 5 and the second equation by -2, we will get c coefficients of 10 and -10.


{{{15b+10c=230}}}
{{{-2b-10c=-22}}}


Adding:
{{{13b+0c=208}}}


{{{13b=208}}}
{{{b=16}}}


Always check your answers.
Remember we said {{{c=-1}}} and {{{b=16}}}.  If that is true, we should be able to substitute these values for the appropriate variables in BOTH of the equations and BOTH of the equations will be true statements.

{{{3b + 2c = 46}}}
{{{3(16)+2(-1) = 48 - 2 = 46}}} TRUE


{{{b + 5c = 11}}}
{{{16 + 5(-1) = 16 - 5 = 11}}} TRUE

Answer checks.