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Question 44237This question is from textbook College Algebra
: Solve each system by substitution. Determine whether each system is independent, inconsistent, or dependent
2y = 1 - 4x
2x + y = 0
This question is from textbook College Algebra
Answer by AnlytcPhil(1806)   (Show Source):
You can put this solution on YOUR website! Solve each system by substitution. Determine whether each system is
independent, inconsistent, or dependent
(a)
x + 2y = 3
2x + y = 5
Solve the first equation for x
x + 2y = 3
x = 3 - 2y
Substitute (3 - 2y) for x in the second equation:
2x + y = 5
2(3 - 2y) + y = 5
6 - 4y + y = 5
6 - 3y = 5
-3y = 5 - 6
-3y = -1
y = 1/3
Now substitute 1/3 for y in
x = 3 - 2y
x = 3 - 2(1/3)
x = 3 - 2/3
Clear of fractions by multiplying thru by LCD = 3
3x = 9 - 2
3x = 7
x = 7/3
This is the most common type of system, when there is exactly
one solution. If you graphed the lines they would cross in
one point (7/3, 1/3). It is called an independent (consistent)
system.
(b)
2y = 1 - 4x
2x + y = 0
Solve the second equation for y
2x + y = 0
y = -2x
Substitute (-2x) for y in the first equation:
2y = 1 - 4x
2(-2x) = 1 - 4x
-4x = 1 - 4x
Add 4x to both sides:
0 = 1
Uh! oh! Zero just doesn't equal to one!
We can't go any further because there are
no variables or unknowns in the equation. But
since we didn't do anything wrong, if we
believed there is any solution to this system,
we'd have to also believe that zero equals one!
But we don't believe that. So we must not
believe this system has a solution either! It
doesn't.
This is a special type of system, when there is
no solution. This is called an inconsistent
system.
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The third type -- you don't have one here -- is when you
get something like 0 = 0 or -1 = -1, when you can't go
any further but the equation is just "a number = itself",
which is true, so the system can't be inconsistent, so it
has infinitely many solutions, and is called a dependent
system.
Edwin
AnlytcPhil@aol.com
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