Answer:
your answer is ‹b ok this is your answer
Answer:
The answer to your question is:
x = 4
y = -1
z = -3
Step-by-step explanation:
3 x + 2 y + z = 7
5 x + 5 y + 4 z = 3
3 x + 2 y + 3 z = 1
![\left[\begin{array}{ccc}3&2&1\\5&5&4\\3&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%261%5C%5C5%265%264%5C%5C3%262%263%5Cend%7Barray%7D%5Cright%5D)
= 45 + 10 + 24 - (30 + 24 + 15)
= 79 - 69
Δ = 10
![\left[\begin{array}{ccc}7&2&1\\3&5&4\\1&2&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D7%262%261%5C%5C3%265%264%5C%5C1%262%263%5Cend%7Barray%7D%5Cright%5D)
= 105 + 6 + 8 - (18 + 56 + 5)
= 119 - 79
Δx = 40
![\left[\begin{array}{ccc}3&7&1\\5&3&4\\3&1&3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%267%261%5C%5C5%263%264%5C%5C3%261%263%5Cend%7Barray%7D%5Cright%5D)
= 27 + 5 + 84 - ( 105 + 12 + 9)
= 116 - 126
Δy = -10
![\left[\begin{array}{ccc}3&2&7\\5&5&3\\3&2&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D3%262%267%5C%5C5%265%263%5C%5C3%262%261%5Cend%7Barray%7D%5Cright%5D)
= 15 + 70 + 18 - (10 + 18 + 105)
= 103 - 133
= -30
Δz = -30
x = Δx /Δ = 40/10 = 4
y = Δy/Δ = -10/10 = -1
z = Δz/Δ = -30/10 = -3
Not necessarily, the more correct definition is opposite reciprocal slopes.
The example used is how horizontal and vertical lines are parallel. Horizontal lines have a slope of 0, also written as 0/1. However, vertical lines have an undefined slope, which isn't necessarily negative. It has a slope of 1/0, which is undefined. In this case, the reciprocal isn't negative.
In all other cases (1 and -1, 2 and -1/2, etc.) yes, the perpendicular pairs are negative and reciprocal.
