How do you solve the following system?: #-8x -6y =3, -2x +3y = -7#

1 Answer
Nov 13, 2015

Answer:

Isolate one variable via adding/subtracting the equations and determine its value, then substitute that back into the equations to determine the other variable.

Explanation:

When confronted with a system of equations that consists of an equal number of variables and equations, it behooves us to attempt to manipulate the equations such that we can isolate and determine the value of one of the variables, which normally allows us to substitute that value for the variable in the equation.

In this case, our system of equations consists of:

A) #-8x -6y = 3#
B) #-2x + 3y = -7#

Now our manipulation begins. How can we add or subtract multiples of #A# and #B# from/to each other, such that we isolate a variable? In order to do this we must cancel out one variable; thus, we must add/subtract the equations such that we either have 0#x# or 0#y#.

From examining the equation, we determine there are two (well, technically 3) ways to do this:

1) #A - 4B#: yields #-8x-6y +8x -12y = 3+28 => -18y = 31 => y = -31/18#
2)#4B - A#: yields #-8x + 12y + 8x + 6y = -28 -3 => 18y = -31 => y = -31/18#
3)#A + 2B#: yields #-8x -6y - 4x + 6y = 3 - 14 => -12x = -11 => x = 11/12#

Note that 1 and 2 are essentially the same operation, simply reordered.

Although above we have already determined the solution (e.g. #x= 11/12, y = -31/18#), it should still be illustrated how one would substitute the determined value for a variable back into the original system in order to obtain the other variable(s).

We might prefer operation #3# here, as it requires the least amount of multiplying. Thus, from operation 3 we determine #x = 11/12#. Now we will substitute #x=11/12# back into equation B to determine #y#. Of note is that we could also insert the value for #x# into equation A, but with equation B we completely cancel the #x# coefficient, which seems marginally easier.

B) #-2x + 3y = -7 => -2(11/12) + 3y = -7 => -11/6 +3y = -7 => 3y = -31/6 => y = -31/18#

The solution we have obtained is #x = 11/12, y = -31/18#

If we wish, then at this point we may substitute the values for #x# and #y# into equation A to ensure they solve both (since we have determined they solve B)

A) #-8x - 6y = 3 => -8(11/12) - 6 (-31/18) = 3 => -22/3 + 31/3 = 3 => 9/3 =3 => 3=3 #

The solution checks out.