Chapter 15. Solving Linear
Equations
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
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= 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
|
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.
3 Systems with More Unknowns
Equations with more than one unknown can be solved if they are
manipulated or massaged into a simpler form. For equations with more than
two or three unknowns, we can obtain complicated formulas for their
solution, but the equations can be solved more efficiently without these
formulas. In this case the method of solution is easier to remember than
the formula. This contrasts with the expectation that I have (accidentally)
built earlier. The shorthand description of calculations that could be
done, that is, formulas for obtaining numbers and quantities, are useful
tools. But we should not insist on using them all the time. Your own
experience or that of others is needed to say when a purely algebraic
approach appears best.
Problem: Find the unknown or forgotten values of
x,y and z when
Solution: Don't panic. From the first equation we see
y = [50/10] = 5. We use this information in the second equation 21 =
2x+3y to get
So 6 = 21-15 = 2x. This gives
2x = 6 or x = 3. Finally, we use the values x = 3 and
y = 5 in the third and last equation14 = x+y+2z
to get 14 = 3+5+2z. Therefore 14 = 8+2z. So 14-8 = 2z. The value of z = [(14-8)/2] = [6/2] = 3. The unknown or forgotten values should
be x = 3, y = 5 and z = 3. Are they correct?
Check:
4 Simplified Problems
Addition-Multiplication Method for non-triangular systems
In the previous problem, we could find the unknowns one at a time. One
method for solving equations is to change, massage and manipulate them into
a form where we can find the unknowns one at a time. We can do this by
adding multiples of equations together.
Example: Solve
Solution: Keep the first equation as is, and add it to the
second. This gives 7 = 2x+y and
The left-hand side 7+5 = 12 and the right-hand side simplifies to
2x+y+2x-y =
4x since adding y and then subtracting it gives the same
result as not doing anything. Therefore the second becomes or is replaced
by
The new second equation suggests x = [12/4] = 3. We use this
value in the first equation to find 7 = 2·3+y. Therefore x =
3 and 7 = 6+y. So the solution should be x = 3 and y =
7-6 = 1. It is easy to check that this proposed
solution satisfies the two equations.
Example: Solve the following system (set) of equations
Solution: To get two equations with equal coefficients for
x, (i) multiply the both sides of the first equation by the number 5
and (ii) multiply both sides of the second equation by 2. This gives a
system of two new equations
Now keep the second equation as is, and subtract it from the first.
This gives a third system
This system has a simple form. We can find y and then
x. Here y = [((60-38))/11] =
[22/11] = 2. We can now use the known value 2 of y to find x.
From 38 = 10x+4y, we may get the value of x in two
ways:
- An arithmetic way: the equation 38 = 10x+4y gives 38 =
10x+4 ·2. So 38 = 10x+8. Therefore 10x =
38-8 = 10. Thus x = [30/10] = 3.
- An algebraic way: the equation 38 = 10x+4y implies or
gives 10x = 38-4y. So x =
[(38-4y)/10] = [(38-4·2)/10] = [30/10] = 3.
Either way the solution is given by x = 3 and y = 2.
Exercise: check this.
5 Examples with Three Unknowns
Example: Solve the following equations (a1), (a2) and (a3).
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(a1
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): 5x
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+2y
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+2z
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=
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+27
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(a2
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): 11x
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+8y
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+2z
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=
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+57
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(a3
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): -3x
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+1y
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-2z
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=
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-16
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A Solution: First subtract equation (a1) from (a2). This implies
yields equation (b2) below.
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(b1
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:) 5x
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+2y
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+2z
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=
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+27
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(b2
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:) 6x
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+6y
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=
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+30
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(b3
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:) -3x
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+1y
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-2z
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=
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-16
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Equations (b1) and (b3) are identical to equations (a1) and (a3),
respectively. We have just changed the labels on them.
Second, add equation (b1) to equation (b3). This gives
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(c1
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): 5x
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+2y
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+2z
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=
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+27
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(c2
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): 6x
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+6y
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=
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+30
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(c3
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): 2x
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+3y
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=
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+11
|
The above steps have eliminated z from the last two equations.
Third, from equation (c2) subtract two times equation (c3). This implies
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(d1
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): 5x
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+2y
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+2z
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=
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+27
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(d2
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): 2x
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|
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=
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+8
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(d3
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): 2x
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+3y
|
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=
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+11
|
Finally, we may change the order of equations. This yields the more
suggestive system of equations:
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(d2
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): 2x
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|
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=
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+8
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(d3
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): 2x
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+3y
|
|
=
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+11
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|
(d1
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): 5x
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+2y
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+2z
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=
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+27
|
This last step was optional. Now we can do the following.
-
- Solve equation (d2) for x. This gives
- Solve equation (d3) for y. This yields
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y =
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1
3
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(11-2x) =
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1
3
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·(11-8) = 1
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|
- Solve equation (d1) for z. This implies
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z =
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1
2
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(27-5x-2y) =
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1
2
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(27-20-2) =
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1
2
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(5) =
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5
2
|
|
|
This in summary yields the solution
respectively.
Exercise: Solve
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(a1
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:) x
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+y
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+ 3z
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=
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+10
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(a2
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:) x
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-y
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+2z
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=
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+5
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(a3
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:) 2x
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+4y
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-5z
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=
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+9
|
Note that you can and should check your answer (values for x,
y and z) satisfy each equation. If one is not satisfied
then there is an error somewhere in your work, the solution or the check.
Exercise: Solve
|
(a1
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:) x
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+y
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+ 3z
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=
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+10
|
|
(a2
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:) x
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-y
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+2z
|
=
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+5
|
|
(a3
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:) 2x
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+4y
|
+5z
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=
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+9
|
See the difference a "small" change in the problem makes.
6 Useful Arithmetic Rules &
Advice
6.1 Useful Arithmetic Rules
In the previous examples we have used rules or properties of
arithmetic to help us in our algebraic manipulations. A short description
of them follows. These rules or properties are all related to the idea that
if two expression are supposed to give the same number when computed, then
using one of them in place of the other in larger expressions should not
change the values given by the larger expressions.
2Talking about the preservation of
equality can be a wordy exercise - to be done at least once, but not
more than thrice.
When a, b and c are shorthand symbols or expressions
representing real numbers, the following properties are useful in solving
equations.
- If a = b, that is if a and b are two
expressions which when evaluated should give the same number (or
quantity) then we should have
ac = bc
a+c = b+c, and
a-c = b-c.
whenever we multiply, add or subtract by another number or
expression c. In brief, equality should be preserved (kept) if
what we do to one side, we also do to the other.2
- If a+b = c+b then a = c.
This follows by subtracting b from both sides.
- If ab = cb and b is nonzero then a =
c.
- If ab = cb and b is nonzero then a =
c. This follows by dividing both sides by b or by
multiplying both sides by [1/(b)] = 1¸b.
- If a = b and c is nonzero then a/c = b/c
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