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ELEMENTARY MATHEMATICS
W W L CHEN and X T DUONG
c
°
W W L Chen, X T Duong and Macquarie University, 1999.
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Chapter 5
POLYNOMIAL EQUATIONS
5.1. Linear Equations
Consider an equation of the type
ax
+
b
=0
,
(1)
where
a, b
∈
R
are constants and
a
6
= 0. To solve such an equation, we Frst subtract
b
from both sides
of the equation to obtain
ax
=
−
b,
(2)
and then divide both sides of this latter equation by
a
to obtain
x
=
−
b
a
.
Occasionally a given linear equation may be a little more complicated than (1) or (2). However, with
the help of some simple algebra, one can reduce the given equation to one of type (1) or type (2).
Example 5.1.1.
Suppose that
4
x
x
+2
=
18
5
.
Multiplying both sides by 5(
x
+ 2), the product of the two denominators, we obtain
20
x
= 18(
x
+2)=18
x
+36
.
Subtracting 18
x
from both sides, we obtain 2
x
= 36, and so
x
=36
/
2 = 18.
†
This chapter was written at Macquarie University in 1999.
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W W L Chen and X T Duong : Elementary Mathematics
Example 5.1.2.
Suppose that
x
−
6
2
+
3
x
4
=
x
+1
.
Multiplying both sides by 4, we obtain 2(
x
−
6) + 3
x
=4(
x
+ 1). Now 2(
x
−
6) + 3
x
=5
x
−
12 and
4(
x
+1)=4
x
+ 4. It follows that 5
x
−
12 = 4
x
+ 4, so that
x
−
16 = 0, giving
x
= 16.
Example 5.1.3.
Suppose that
6
x
−
1
4
x
+3
=
3
x
−
7
2
x
−
5
.
Multiplying both sides by (4
x
+ 3)(2
x
−
5), the product of the two denominators, we obtain
(6
x
−
1)(2
x
−
5) = (3
x
−
7)(4
x
+3)
.
Now (6
x
−
1)(2
x
−
5) = 12
x
2
−
32
x
+ 5 and (3
x
−
7)(4
x
+3)=12
x
2
−
19
x
−
21. It follows that
12
x
2
−
32
x
+5=12
x
2
−
19
x
−
21, so that
−
13
x
+ 26 = 0, whence
x
=2.
Example 5.1.4.
Suppose that
3
x
2
x
=
3
x
−
3
2
x
+5
.
Multiplying both sides by (2
x
+ 3)(2
x
+ 5), the product of the two denominators, we obtain
(3
x
+ 1)(2
x
+5)=(3
x
−
3)(2
x
.
Now (3
x
+ 1)(2
x
+5)=6
x
2
+17
x
+ 5 and (3
x
−
3)(2
x
+3)=6
x
2
x
−
9. Check that the original
equation is equivalent to 14
x
+ 14 = 0, so that
x
=
−
1.
We next consider a pair of simultaneous linear equations in two variables, of the type
a
1
x
+
b
1
y
=
c
1
,
a
2
x
+
b
2
y
=
c
2
,
(3)
where
a
1
,a
2
,b
1
2
,c
1
2
∈
R
. Multiplying the Frst equation in (3) by
b
2
and multiplying the second
equation in (3) by
b
1
, we obtain
a
1
b
2
x
+
b
1
b
2
y
=
c
1
b
2
,
a
2
b
1
x
+
b
1
b
2
y
=
c
2
b
1
.
(4)
Subtracting the second equation in (4) from the Frst equation, we obtain
(
a
1
b
2
x
+
b
1
b
2
y
)
−
(
a
2
b
1
x
+
b
1
b
2
y
)=
c
1
b
2
−
c
2
b
1
.
Some simple algebra leads to
(
a
1
b
2
−
a
2
b
1
)
x
=
c
1
b
2
−
c
2
b
1
.
(5)
On the other hand, multiplying the Frst equation in (3) by
a
2
and multiplying the second equation in
(3) by
a
1
, we obtain
a
1
a
2
x
+
b
1
a
2
y
=
c
1
a
2
,
a
1
a
2
x
+
b
2
a
1
y
=
c
2
a
1
.
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