MAT1503 ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 20
University of South Africa
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MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022)
- Exam (elaborations) • 14 pages • 2022
- R252,21
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MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022)
(a) (i) We shall do this by substitution. Thus
2 (t + 2) + 4 (−2t − 1) + 6t
= 2t + 4 − 8t − 4 + 6t
= 0
Also
4 (t + 2) + 5 (−2t − 1) + 6t
= 4t + 8 − 10t − 5 + 6t
= 3
and finally
7 (t + 2) + 8 (−2t − 1) + 9t
= 7t + 14 − 16t − 8 + 9t
= 6
Thus, x = t+2, y = −2−1, z = t satisfy all three equation. This solution is general since it has the parameter t, enabling the generation of any other solution from this...
Exam (elaborations)
MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022)
Last document update:
ago
MAT1503 - ASSIGNMENT 03 SOLUTIONS (SEMESTER 01 -2022) (a) (i) We shall do this by substitution. Thus 2 (t + 2) + 4 (−2t − 1) + 6t = 2t + 4 − 8t − 4 + 6t = 0 Also 4 (t + 2) + 5 (−2t − 1) + 6t = 4t + 8 − 10t − 5 + 6t = 3 and finally 7 (t + 2) + 8 (−2t − 1) + 9t = 7t + 14 − 16t − 8 + 9t = 6 Thus, x = t+2, y = −2−1, z = t satisfy all three equation. This solution is general since it has the parameter t, enabling the generation of any other solution from this...
R252,21
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