In an algebraic sentence, the verb is typically the equal
sign = .
That sentence -- that statement -- will logically imply
other statements. Let us follow the logical sequence that leads to
the final statement, which is the solution.
Thus we solve an equation by logically transforming it --
changing its form -- statement by statement, line by line, until x
finally is isolated on the left. That is how books on mathematics
are written (but unfortunately, not books that teach algebra!). Each
line is its own readable statement that follows from the line above --
with no crossings out!
|
(1) |
abcx − d + e − f |
= |
0 |
|
| implies
|
(2) |
abcx |
= |
d −
e + f |
|
| implies
|
(3) |
x |
=
|
d −
e + f.
abc |
Transpose the terms first. Line
(2).
It is not
necessary to write the term 0 on the right.
Problem
2.
Write the logical sequence of statements that
will solve this equation for
x :
|
(1) |
2x + 5 |
= |
27 |
|
| implies
|
(2) |
2x |
= |
27 − 5 =
22 |
|
| implies
|
(3) |
x |
=
|
22
2 |
|
| implies
|
(4) |
x |
=
|
11. |
Problem
3.
Solve for
x : (
p −
q)
x +
r =
s
Problem
4.
Solve for
x :
ab(
c
+
d)
x −
e +
f = 0
Problem
5.
Solve for
x : 2
x + 1= 0
x = −½
Problem
6.
Solve for
x :
ax = 0
(
a
0).
Now, when the product of two numbers is 0, then at least one of them
must be 0. Therefore, any equation with that form has the
solution, x = 0.
| We could solve that formally, of course, by dividing
by a. |
0
a |
=0. |
Problem
7.
Solve for
x :
| 4x − 2 |
= |
−2 |
|
| 4x |
= |
−2 + 2 = 0 |
|
| x |
= |
0 |
Problem
8.
Write the sequence of statements that will
solve this equation:
| (1) |
6 − x |
= |
9 |
|
| (2) |
−x |
= |
9 − 6 |
|
| (3) |
−x |
= |
3 |
|
| (4) |
x |
=
|
−3 |
An equation is
"solved" when positive x is isolated on the left. We go from
line (3) to line (4) by changing
the signs on both sides.
Alternatively, then, we could have
changed all the signs immediately:
| (1) |
6 − x |
= |
9 |
|
| (2) |
−6 +
x |
= |
−9 |
|
| (3) |
x |
= |
−9 + 6 =
−3 |
Problem 9.Solve for x
: 3 − x |
= |
−5 |
|
|
x |
= |
8 |
Problem
10.
Solve for
x :
| 5 − 2x |
= |
−11 |
|
| −2x |
= |
−11 − 5 |
|
| 2x |
= |
16 |
|
| x |
= |
8 |
Transposing versus exchanging sides
We can easily solve this -- in one line -- simply by
transposing x to the left and what is on the left, to the
right:
x = c −a − b.
In this Example, +x is on the right. Since we
want +x on the left, we can achieve that by exchanging
sides:
c + x = a +
b
Note: When we exchange sides, no signs
change.
The solution easily follows:
c + x = a +
b − c
In summary, when −x is on the right, it is skillful
simply to transpose it. But when +x is on the right, we may
exchange the sides.
Problem
11.
Solve for
x :
|
p + q |
= |
r − x − s |
|
| Transpose: |
|
|
|
x |
= |
r −
s − p − q |
Problem
12.
Solve for
x :
|
p − q + r |
= |
s + x |
|
| Exchange sides: |
|
|
|
s + x |
= |
p −
q + r |
|
|
x |
= |
p −
q + r − s |
Problem
13.
Solve for
x :
| 0 |
= |
px + q |
|
| px + q |
= |
0 |
|
|
| px |
= |
− |
q |
|
| x |
= |
− |
q
p |
Problem
14.
Solve for
x :
| −2 |
= |
−5x + 1 |
|
| 5x |
= |
1 + 2 =
3 |
|
| x |
= |
3
5 |
Here are the basic theorems:
Theorem 1. |
|
If |
|
|
|
x + a |
= |
b, |
|
| then |
|
|
|
|
x |
= |
b − a. |
For, upon adding −a to both
sides:
| x + a + (−a) |
= |
b + (−a) |
|
| x + 0 |
= |
b
− a |
|
| x |
= |
b − a. |
Which is what we wanted to prove.
Having proved this, then from now on we may write
immediately:
|
x + a |
= |
b |
|
| implies |
x |
= |
b − a. |
That is, we may transpose.
Similarly, by adding a to both
sides, we could prove:
| If |
|
|
|
|
x − a |
= |
b, |
|
| then |
|
|
|
|
|
x |
= |
b + a. |
Theorem 2. |
|
If |
|
|
|
ax |
= |
b, |
|
| then |
|
|
|
|
x |
= |
b
a |
. |
| For, on multiplying both sides by |
1
a |
: |
1
a |
·ax |
= |
1
a |
·b |
|
|
1·x |
= |
b
a
| (Definition
of division) |
|
|
x |
= |
b
a
| . |
Which is what we wanted to prove.
Having proved this, then from now on we may write
immediately:
Similarly, by multiplying both sides by a, we could
prove:
