*This preview shows
page 1. Sign up
to
view the full content.*

**Unformatted text preview: **x-axis of length 6 and a minor axis along the y -axis of length 4. We see from
the graph that the two curves intersect at their y -intercepts only, (0, ±2).
2. We proceed as before to eliminate one of the variables
(E 1)
x2 + y 2 = 4
(E 2) 4x2 − 9y 2 = 36 Replace E 2 with −− − − −→
−−−−−
−4E 1 + E 2 (E 1) x2 + y 2 = 4
(E 2) −13y 2 = 20 8.7 Systems of Non-Linear Equations and Inequalities 533 Since the equation −13y 2 = 20 admits no real solution, the system is inconsistent. To verify
this graphically, we note that x2 + y 2 = 4 is the same circle as before, but when writing the
2
2
second equation in standard form, x − y4 = 1, we ﬁnd a hyperbola centered at (0, 0) opening
9
to the left and right with a transverse axis of length 6 and a conjugate axis of length 4. We
see that the circle and the hyperbola have no points in common. y y 1 −3 −2 −1 1 1 2 3 −3 x −2 −1 −1 x2 + y 2 = 4
4x2 + 9y 2 = 36 Graphs for 1 2 3 x −1 Graphs for x2 + y 2 = 4
4x2 − 9y 2 = 36 3. Since there are n...

View
Full
Document