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Note that this just produces three parametric equations:
In the
plane, this is just he parametric equation for an ellipse (as a function of u). The z is simply a linear function.
The surface is then an ellipse extruded along the z-axis. We get a elliptic cylinder.
In the
The surface is then an ellipse extruded along the z-axis. We get a elliptic cylinder.
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.
