
Bradford T. answered 01/11/21
Retired Engineer / Upper level math instructor
If you plot all three linear equations, they all intersect at (6,-2)
You can also solve by subtracting the third equation from the second, leaving
5y = -10 or y = -2 and substituting y = -2 into the second equation gives
2x+6 = 18 --> x = 6
Substituting both x and y into the first equation verifies the result
6 - 2 = 4
You will get the same result if you add 3 times the first equation to the second:
5x = 30 --> x = 6
and substituting back into the second equation
12 -3y = 18 --> -3y = 6 ==> y = -2
Using matrices, you can show that one equation is redundant
|1 1 4| r3 = r3-r2 |1 4 4| r2 = r2-2r1 | 1 1 4|
|2 -3 18| |2 -3 18| | 0 -5 10|
|2 -8 28| |0 -5 10| | 0 -5 10|
r3 = r3 -r2
| 1 1 4 | r2 = r2/-5 | 1 1 4| r1 = r1 - r2 |1 0 6|
| 0 -5 10| | 0 1 -2| |0 1 -2|
| 0 0 0| | 0 0 0| |0 0 0|
x = 6
y = -2