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MODULE 1
1. REVIEW OF SOME BASIC ALGEBRA
(1)
Solving Equations
You should be able to solve for
x
:
a
+
1
b
=
c
−
a
d
+
e
x
+
c
and get
x
=
e
(
ba
+1)
b
(
c
−
a
)
−
d
(
ba
−
c
Common mistakes and strategies:
.
a
b
+
c
6
=
a
b
+
a
c
, but
a
+
b
c
=
a
c
+
b
c
[You can’t separate a demominator, but you can divide a single denominator into each numerator term]
.
1
y
=
a
+
1
a
+
b
implies
y
=
1
a
+
1
a
+
b
=
a
+
b
a
(
a
+
b
)+1
=
a
+
b
a
2
+
ab
+1
[The Frst step is by inverting both sides, being careful to invert each side as a whole; the second step is by
multiplying numerator and denominator by (
a
+
b
)]
Note that combining di±erent algebraic fractions can usually only be done by multiplying numerator and
denominator of
each
fraction in such a way as to
create
a common denominator:
1
a
+
1
a
+
b
+
1
c
=
1
a
·
(
a
+
b
)
c
(
a
+
b
)
c
+
1
a
+
b
·
ac
ac
+
1
c
·
a
(
a
+
b
)
a
(
a
+
b
)
=
(
a
+
b
)
c
+
ac
+
a
(
a
+
b
)
a
(
a
+
b
)
c
=
2
ac
+
bc
+
a
2
+
ab
a
2
c
+
abc
Similarly, a fraction in the denominator can be simpliFed by multiplying entire numerator and entire de
nominator by common denominator of the fractions in the denominator:
a
+
c
1
a
+
1
a
+
b
=
a
+
c
1
a
+
1
a
+
b
·
a
(
a
+
b
)
a
(
a
+
b
)
=
a
(
a
+
b
)
1
a
·
a
(
a
+
b
)+
1
a
+
b
·
a
(
a
+
b
)
=
a
2
+
ab
(
a
+
b
a
=
a
2
+
ab
2
a
+
b
(2)
Quadratic equations
The standard equations studied in high school algebra in two dimensions involving quadratic powers of the
variables are the following. In these a constant may be written as
a
2
or
b
2
to indicate it is positive, since
squares of real numbers are always positive. Constants written as
a
or
b
may be either positive or negative.
.
a
2
x
2
+
a
2
y
2
=
c
2
(circle)
.
a
2
x
2
+
b
2
y
2
=
c
2
(ellipse)
.
½
a
2
x
2
−
b
2
y
2
=
c
2
b
2
y
2
−
a
2
x
2
=
c
2
(hyperbola)
.
½
y
=
ax
2
+
bx
+
c
x
=
ay
2
+
by
+
c
(parabola)
.
xy
=
c
(hyperbola)
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View Full DocumentThe scientist or engineer should be able to name and sketch these equations on sight. This is most eﬃciently
done by knowing the possible shapes and then checking the easy cases
x
=0and
y
= 0.
For example,
consider the ellipse 9
x
2
+
y
2
= 9. An ellipse is always “eggshaped” but is it elongated along the
x
axis
or the
y
axis? Consider
x
= 0. Then
y
2
=9so
y
=
±
3, and the ellipse goes through the points
±
3on
the
y
axis. Consider
y
= 0. Then 9
x
2
=9
,so
x
=
±
1 and the ellipse goes through the points
±
1onthe
x
axis. Thus the ellipse is elongated along the
y
axis. Its graph is given in plot A below. Note that
x
=0
corresponds to points on the
y
axis and
y
= 0 corresponds to points on the
x
axis. This is an observation
which sometimes confuses students.
Consider 6
x
2
−
y
2
= 1. It is clearly a hyperbola, but along which axis is it oriented? Consider
x
= 0. Then
y
2
=
−
1 which is never true for any real number, so
x
= 0 can never satisfy the equation. This already tells
you that no part of the
y
axis is on the curve; ie., the hyperbola is oriented along the
y
axis. Consider
y
=0.
Then 6
x
2
= 1, so the hyperbola goes through the points
±
1
√
6
on the
x
axis. The graph is given in plot B
below. The graph of 6
y
2
−
x
2
= 1 may be sketched by similar reasoning, and is given in plot C below.
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 Spring '08
 DONTREMEMBER
 Algebra, Geometry, Equations

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