Consistent and Inconsistent Systems

Jan 2020
2
0
United States
Hi everyone,

I am starting to learn Linear Algebra and came across an example in my textbook that explains consistent and inconsistent systems.

The below examples have been provided (numbers on the right of 'x' are meant to be subscripted):

a)
x1 + x2 = 2
x1 - x2 = 2

b)
x1 + x2 = 2
x1 + x2 = 1

c)
x1 + x2 = 2
-x1 - x2 = -2

I understand why the system in a) is consistent with one solution, and why the system in b) is inconsistent.

However, I do not understand why the system in 3 is consistent with infinitely many solutions and not inconsistent (parallel), especially as they are intersecting the x-axis at two different points, 2 and -2. If I were to try and fit logic to come to the same answer, I know that multiplying one of the equations by -1 would result in both of the equations being the same, but I do not understand why I would do that.

The book has also provided graphs showcasing systems a), b) and c), which I am attaching here. I understand why the first equation of system c) would look like this, but why wouldn't the second system of c) reflect a parallel line to this?

Any help in explaining would be greatly appreciated!
 

skeeter

Math Team
Jul 2011
3,206
1,731
Texas
c)
x1 + x2 = 2
-x1 - x2 = -2

I understand why the system in a) is consistent with one solution, and why the system in b) is inconsistent.

However, I do not understand why the system in 3 is consistent with infinitely many solutions and not inconsistent (parallel), especially as they are intersecting the x-axis at two different points, 2 and -2.
No, they do not intersect at different values ... setting $x_1=0$ in both equations yields $x_2 = 2$ in both


They are the same line. Multiply every term in either equation by $-1$, you'll get the other equation.

another way to look at it ...

$x_1+x_2=2$

add $-x_1-x_2$ to both sides ...

$0 = 2 - x_1 - x_2$

add $-2$ to both sides ...

$-2 = -x_1-x_2$
 
Jan 2020
2
0
United States
No, they do not intersect at different values ... setting $x_1=0$ in both equations yields $x_2 = 2$ in both


They are the same line. Multiply every term in either equation by $-1$, you'll get the other equation.

another way to look at it ...

$x_1+x_2=2$

add $-x_1-x_2$ to both sides ...

$0 = 2 - x_1 - x_2$

add $-2$ to both sides ...

$-2 = -x_1-x_2$
Thanks, Skeeter! Apologies, looking back, in context of your response, that was a bit of a silly question. Appreciate the reply!