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Unformatted text preview: Solutions to Quiz 12A
www.math.uﬂ.edu/˜harringt
October 31, 2007
1. Find the solution set for each system.
1
x
2
1
x
4 + 1y = 3
3
− 2 y = −1
3 1
x
2
1
x
4 (a) + 1y = 3
3
⇒
− 2 y = −1
3 1
6( 1 x + 3 y )
= 6(3)
2
⇒
2
−12( 1 x − 3 y ) = −12(−1)
4 3x + 2y
= 18
−3x + 8y = 12 Next, if we use elimination, we have the following:
3x + 2y
−3x + 8y
−−−−−
10y =
=
−
= 18
12
−−
30 Thus, y = 3. To solve for x, we can plug y = 3 into 3x + 2y = 18.
So we have:
3x + 2(3) = 18
3x = 12
x=4
So the solution set is {(4, 3)}.
(b) 3
x
2 − 1y = 7
2
2
9x − 3y = 21
− 1y = 7
2
2
⇒
9x − 3y = 21
3
x
2 3
7
2( 2 x − 1 y ) = 2( 2 )
2
⇒
9x − 3y
= 21 1 3x − y = 7
9x − 3y = 21 So our system can be reduced to:
9x − 3y = 21
9x − 3y = 21
By using elimination, we have that 0 = 0 so that the system is
dependant. So we have inﬁnitely many solutions! (We can not
write each individual solution so we must use set notation.) Our solution set is {(x, y ) : 9x − 3y = 21}.
2. If a system of equations has no solution it is said to be inconsistant. 2 ...
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This note was uploaded on 07/12/2011 for the course MAC 1140 taught by Professor Williamson during the Fall '08 term at University of Florida.
 Fall '08
 WILLIAMSON
 Math, Calculus, Algebra

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