Evaluate x(–y + z) for x = 3, y = 3, and z = 1. –8 –6 12 10
1 answer:
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Answer:
(x, y) = (2, 5)
Step-by-step explanation:
I find it easier to solve equations like this by solving for x' = 1/x and y' = 1/y. The equations then become ...
3x' -y' = 13/10
x' +2y' = 9/10
Adding twice the first equation to the second, we get ...
2(3x' -y') +(x' +2y') = 2(13/10) +(9/10)
7x' = 35/10 . . . . . . simplify
x' = 5/10 = 1/2 . . . . divide by 7
Using the first equation to find y', we have ...
y' = 3x' -13/10 = 3(5/10) -13/10 = 2/10 = 1/5
So, the solution is ...
x = 1/x' = 1/(1/2) = 2
y = 1/y' = 1/(1/5) = 5
(x, y) = (2, 5)
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The attached graph shows the original equations. There are two points of intersection of the curves, one at (0, 0). Of course, both equations are undefined at that point, so each graph will have a "hole" there.
Answer:
Solution : Parabola
Step-by-step explanation:
As you can see only one variable is square in this situation, so it can only be a parabola. We can prove that it is a parabola however by converting it into standard form (x - h)^2 + (y - k)^2.
Respectively it's properties would be as follows,
