Without graphing , how do you decide whether the following system of linear equations has one solution , infinitely many solutions or no solution?

1 Answer
Mar 31, 2016

Answer:

A system of #N# linear equations with #N# unknown variables that contains no linear dependency between equations (in other words, its determinant is non-zero) will have one and only one solution.

Explanation:

Let's consider a system of two linear equations with two unknown variables:
#Ax+By=C#
#Dx+Ey=F#

If pair #(A,B)# is not proportional to pair #(D,E)# (that is, there is no such number #k# that #D=kA# and #E=kB#, which can be checked by condition #A*E-B*D != 0#) then there is one and only one solution:

#x=(C*E-B*F)/(A*E-B*D)#, #y=(A*F-C*D)/(A*E-B*D)#

Example:
#x+y=3#
#x-2y=-3#

Solution:
#x=(3*(-2)-1*(-3))/(1*(-2)-1*1)=1#
#y=(1*(-3)-3*1)/(1*(-2)-1*1)=2#

If pair #(A,B)# is proportional to pair #(D,E)# (which means that there is such number #k# that #D=kA# and #E=kB#, which can be checked by a condition #A*E-B*D = 0#), there are two cases:

(a) infinite number of solutions if #C# and #F# are proportional with the same coefficient as #A# and #D#, that is #F=kC#, where #k# is the same coefficient of proportionality;
Example:
#x+y=3#
#2x+2y=6#
Here #k=2# for all pairs: #D=2A#, #E=2B#, #F=2C#.
The second equation is a trivial consequence of the first (just multiply the first equation by #2#) and, therefore, provides no additional information about unknown, reducing the number of equations, effectively, to 1.

(b) no solutions at all, if #F!=kC#
Example:
#x+4y=3#
#2x+8y=5#
In this case equations contradict each other since, by multiplying the first by 2, we derive to equation #2x+8y=6#, which cannot have common solution with #2x+8y=5# since the left parts of these two equations are equal, but the right parts are not.