Solving Linear Equations in One Unknown -Numerically and Algebraically
From Chapter 15, Volume 2, Three Skills for Algebra
Here are some more examples in which we solve equations. Our aim is to
become familiar or at ease with handling and manipulating equations. So
we look at the algebraic solution of equations containing one or more
unknown numbers.
1 One Unknown
1.1 First Example
When we let x = 5, we have 2x = 10 and 4x
¹ 15. Suppose now we forgot the value of
x which made 2x = 10, could we find the value of x
from the equation 2x = 10? The answer is yes. We can solve for the
unknown or forgotten value of x as follows:
In this solution, we used the property [(ab)/(b)] =
b with the role of a played by x and the role of
b played by 2. This gives the first equality. The second equality
follows from assumption that 2x = 10. The latter allows 2x to
be replaced by its value 10. Another way to look at this solution is to say
Therefore
Hence
The manipulation process here creates new equalities from previous
ones until an expression
appears. How we get find the value of x from an equation
involving x or other unknowns is a matter of taste.1
1.2 Second Example
Problem: Find the value of x which satisfies the equation
7x+9 = 65.
Solution: The aim is to manipulate (or change or massage) the
given equation
to get a new one of the form
The first step is to subtract 9 from both sides. This gives
Some of you may know that 65-9 = 56. We
could write 56 instead of 65-9. A next step to
further isolate x is to divide by 7 (or multiply by [1/7]) since
[(7x)/7] = x. This manipulation gives
Therefore
The isolation of x is complete. The solution is x =
8. To check this, just in case we made a mistake, observe when x =
8, we have 7x+9 = 7·8+9 = 56+9 = 65. So the original the equation
7x+9 = 65 holds (is satisfied, is true) when x = 8.
1.3 Third Example
Problem: Find the value of x which satisfies 5x+6
= 117.
Solution: The aim is to manipulate the given equation
to get a new one of the form
A first step is to subtract 6 from both sides. This gives
A next step to further isolate x is to divide by 5 (or
multiply by [1/5]) since [(5x)/5] = x. This manipulation
gives
Therefore
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x =
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111
5
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= 22+
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1
5
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= 22
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1
5
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The isolation of x is done. The solution is x =
[111/5]. To check this, just in case we made a mistake, observe when
x = [111/5], we get 5x+6 = 111+6 = 117.
The solution
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x =
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111
5
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=
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111 ×2
5 ×2
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=
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222
10
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= 22.2 = 22
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1
5
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can be written in several ways. Which way we prefer is a matter of
taste.
Here are some more examples in which we solve equations. Our aim is to
become familiar or at ease with handling and manipulating equations. So
we look at the algebraic solution of equations containing one or more
unknown numbers.
2 Algebraic Shorthand Solution
In a play or movie, the roles are more important than the actors, stars
excepted. That is, any role can be played by any actor. But after the
cast is selected, each role is usually played by only one actor, and each
actor usually plays only one role. Once a play (or scene) is finished,
the actors can take roles in another play (or scene). Likewise, in
algebra, we have choice in the selection of the shorthand notation in
which a problem or its solution is posed. But after the selection, the
choice should be fixed at least temporarily. Once the problem and
solution have been treated, the shorthand in it can be recycled in
another plot.
2.1 Third Example Revisited
The role of x in the third example can be played by any other
letter, for instance y. We will repeat the third example problem
with y in place of x. (This is mathematics ad nauseum.)
Problem: Find the value of y which
satisfies 5y+6 = 117. (This problem is identical to the previous
one, except the shorthand symbol for the forgotten or unknown number is
now the letter y instead of the letter x. The solution is
identical. It is given or repeated next. Excuse the repetition, but you
must see that it is a repetition.)
Solution: The aim is to manipulate the given
equation
to get a new one of the form
A first step is to subtract 6 from both sides.
This gives
A next step to further isolate y is to
divide by 5 (or multiply by [1/5]) since [(5y)/5] = y. This
manipulation gives
Therefore
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y =
|
111
5
|
= 22+
|
1
5
|
= 22
|
1
5
|
|
|
The isolation of y is done. The solution is y =
[111/5]. To check this, just in case we made a mistake, observe when
y = [111/5], we get 5y+6 = 111+6 = 117.
2.2 An Algebraic Pattern
Each of the above examples has the form ax+b = c in
which the numbers a, b and c are given, and x
is initially unknown. In the first example, the roles of a,
b and c were played or given by 7, 9 and 65. That gave the
equation 7x+9 = 65. In the second example 5x+6 = 117, the
number 5 is used in place of a, the number 6 plays the role of
b and the number 117 is given by c.
General Problem: Find x if ax+b = c.
ALGEBRAIC SHORTHAND SOLUTION. We follow the pattern set in the previous
examples. First we subtract b from both sides of the equation
ax+b = c. This gives
Next, we observe if a is nonzero,
Thus the formula for x is
This gives a recipe for x no matter what values of a,
b and c are given in the problem: find x if
ax+b = c. The formula can be used when a
¹ 0. Division by zero is not permitted or done
in arithmetic. It is not possible.
Check: When x = [(c-b)/(a)], we see ax+b =
a·[(c-b)/(a)] =
(c-b)+b = c as
hoped.
The recipe
describes and gives the solution to many problems of the form
ax+b = c.
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Problem
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Solution
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ax+b = c
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x = [(c-b)/(a)]
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5x+6 = 65
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x = [(65-6)/5]
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7x+9 = 117
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x = [(117-9)/7]
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7y+9 = 117
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y = [(117-9)/7]
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123x+456 = 12067
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x = [(12067-456)/123]
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100x+(-20) = 800
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x = [(800-(-20))/100] = [(800+20)/100] = 8.2
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100x-20 = 800
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x = [(800-(-20))/100] = [(800+20)/100] = 8.2
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[4/5]x+4 = 10.2
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x = [(10.2-4)/([4/5])]
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3z+7 = 19
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z = [(19-7)/3]
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The formula x = [(c-b)/(a)] describes and gives a solution to
many problems of the form ax+b = c. We can further
use this recipe without repeating each time, the reasoning that led to
it.
EXTRA. The above formula for x can be used to solve the equation
ax-d = c by putting
b = -d. The equation
ax-d = c can be rewritten
as ax+(-d) = c since
subtraction of d can be replaced by the addition of the number
-d.
Here are some more examples in which we solve equations. Our aim is to
become familiar or at ease with handling and manipulating equations. So
we look at the algebraic solution of equations containing one or more
unknown numbers.
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