How do solve the following linear system?: # x+2y=5 , -6x+3y=-12 #?

I apologize in advance for the length of this explanation, but it's good to know multiple ways of doing something (:

Substitution: When we use the substitution method to solve linear systems, we want to isolate one of the variables in one of the equations and use its represented value to "plug in", or substitute, into the second equation. Let's isolate #x# from the first equation of this system, since it has a coefficient of #1#:

#x + 2y = 5# Given
#x = -2y + 5# Subtract #(2y)# from "both sides".

Now that we know what #x# is, we can substitute this value (#-2y + 5#) into the second equation to solve for #y# (for every #x# that you see, write (#-2y + 5#) instead):

#-6x + 3y = -12# Given
#-6(-2y + 5) + 3y = -12# Substitute
#12y - 30 + 3y = -12# Distribute
#15y = 18# Combine like terms
#y = 18/15 = 6/5# Isolate #y# + Simplify

This means that the y-coordinate of the lines' intersection point is #6/5#, or #1.2#. To find the x-coordinate of the point, all we need to do is substitute this numerical y-value into either equation and solve for #x#, OR we can simply use our value for #x# that we found earlier: #x = -2y + 5#:

#x = -2y + 5# Given
#x = -2(6/5) + 5# Substitute
#x = (-12/5) + 25/5# Multiply + Convert
#x = 13/5# Add with the like denominators

And here is our x-coordinate of the intersection point; #13/5#, or #2.6#. This means that the solution to this linear system is #(13/5, 6/5)#, or #(2.6, 1.2)#.

Making Zeros: This method is most useful when the coefficients in your system are relatively small. We can manipulate each equation to isolate a single common variable by multiplying them by respective values to cancel out the variable that we will find second:

#x + 2y = 5# Given Equation 1
#-6x + 3y = -12# Given Equation 2

#6x + 12y = 30# Multiply Equation 1 by (6)
#-6x + 3y = -12# Vertically align Equation 2 for Addition

#0x + 15y = 18# Add like terms vertically
#15y = 18# Simplify
#y = 18/15 = 6/5# Isolate #y#

Here we have our y-coordinate of the solution. You'll notice we got the same value as when we used the substitution method. To find the x-coordinate of the solution point, we can do the same thing as the substitution method and plug in #y# for either equation:

#x + 2y = 5# Given
#x + 2(6/5) = 5# Substitute
#x + 12/5 = 25/5# Multiply + Convert
#x = 25/5 - 12/5# Isolate #x#
#x = 13/5# Simplify

As you can see, this method yields the same solution point for the system, #(13/5, 6/5)#.

Graphing: This method is exactly what it sounds like--we graph! So, graph both of the lines however you like (using a table to plug in values, converting to slope-intercept form first, etc.). The point where they intersect is the solution to the system; it should be around #(13/5, 6/5)#.

PLEASE NOTE: Graphing is the least accurate method of these three (unless using a graphing utility, of course), so use only when necessary.