Paul H. answered • 07/30/23
I have taught your college or university course.
The statement of this problem is slightly unusual, but not that much. i'm guessing the statement is throwing you somewhat.
4x+16y = 36-6x+8y = 26 actually gives us 3 equations: We take two of the three sides set equal, in turn.
4x+16y = 26 [1]
36-6x+8y = 26 [2]
4x+16y = 36-6x+8y [3]
which can be written into a more standard form by rearranging and simplifying (combining) like terms:
4x+16y=26 [1]
-6x+ 8y=-10 [2]
10x+ 8y=36 [3]
Comments:
We have now a system of three equations in two unknowns, x and y.
The possibilities for solution of any linear system are NO SOLUTION, A UNIQUE SOLUTION, an INFINITE NUMBER OF SOLUTIONS. We need to determine which of these holds. We need further analysis to know.
First, note that not every possibility is a solution. For instance, if x=1 and y=1, that does not work. That would make 20=26 in [1].
Some problems have an infinite number of solutions, where one variable determines the value of the other. But if x=1, then 4(1)+16y=26 would make y=22/16 in [1] and -6(1)+8y=-10 would make y=-4/8 in [2] and neither works in [3].
The best way to continue is to solve the system of 3 equations by some form of the Gauss-Jordan method, done in the usual manner.
For instance, the initial matrix
[ 4 16 26
-6 8 -10
10 8 36 ]
reduces by standard row-operation steps to the Row Reduced Echelon Form
[ 1 0 23/8
0 1 29/32
0 0 0 ]
This latter gives x=23/8 and y=29/32 in the solution, which checks in all 3 equations.
Note that the last row in the RREF matrix simply says 0x+0y=0 which is true (consistent) so does not affect the solution given by the first two rows. Final result is a UNIQUE SOLUTION (x,y) = (23/8,29/32).
