Question #11a35

When we are given a system of equations with multiple variables, our goal is to find what values for those variables could make all of the equations in our system true.

In this case, we are given the system

#{(y = -6),(y=5x+14):}#

So we want to know what values of #x# and #y# make both of those equations true. The first equation gives us that #y# must be #-6#, meaning all that remains is to solve for #x#. We will do so using substitution.

The substitution property of equality means that if two things are equal, then wherever one appears, we can switch it out for the other.

In this case, we have #y = -6#. This means that wherever #y# appears, we can substitute #-6#. If we do so in the second equation, that gives us

#color(red)(y) = 5x + 14#

#color(red)(-6) = 5x+14#

Now we can use algebraic manipulation to solve for #x#.

#-6=5x+14#

#=> -6-14 = 5x+14-14#

#=> -20 = 5x#

#=> (-20)/5 = (5x)/5#

#=> -4 = x#

Thus, the solution to the system is #(x, y) = (-4, -6)#

Because we already have both equations solved for #y#, we can set the two expressions which are both equal to #y# equal to each other and solve for #x#. This process is called substitution, in the case of solving linear systems.

Note: 'solved for #y#' simply means that #y# is isolated on one side of the equals sign and has only the understood #+1# coefficient and the understood #+1# exponent.

So, let's look at this system of equations and how to find its solution.

#y = -6#
#y = 5x + 14#

As I stated earlier, set the expressions for #y# equal to each other and use this new equation to find the value of #x#.

#-6 = 5x + 14#
#-6 - 14 = 5x + 14 - 14#
#-20 = 5x#

#(-20)/5 = (5x)/5#

#-4 = x#

Since we know from the original system of equations that #y = -6#, we now have both coordinates of the point of intersection of the two lines, which is the solution of the system. The solution is the point #(-4, -6)#.