The following system has one solution: x = 2, y = β2, and z = 3.
4x β 2y + 5z = 27 Equation 1
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Mathematics, 31.05.2021 01:00, mathater15
The following system has one solution: x = 2, y = β2, and z = 3.
4x β 2y + 5z = 27 Equation 1
x + y = 0 Equation 2
βx β 3y + 2z = 10 Equation 3
1. Solve the system provided by Equations 1 and 3. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, y, and z in terms of the parameter t.)
(x, y, z) =
2. Solve the system provided by Equations 2 and 3. (If there is no solution, enter NO SOLUTION. If the system has an infinite number of solutions, express x, y, and z in terms of the parameter t.)
(x, y, z) =
3. How many solutions does each of these systems have?
Equations 1 and 2:
(a)zero
(b) one
(c) infinitely many
Equations 1 and 3:
(a)zero
(b)one
(c)infinitely many
Equations 2 and 3:
(a)zero
(b)one
(c) infinitely many
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Answers: 2
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