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"*fhwi E‘” ‘6’? on Adgebra (Algebraic rearrangements) 7 “KW, immwﬂ .___.____: ..... Often, we can convert a complex problem into algebraic symbols and
an equation. Then, we can use the rules of algebra to solve the equation.
Also, most physical laws can be put in their most simple form as a
mathematical equation. Once in equation form, it can be algebraically
rearranged to show other aspects of the principle. Unfortunatdy. many
students are not familiar with the methods and use of algebra. If you
have not taken algebra in several yams, you will need to carefully review
the guidelines below. This chemisu'y course will use algebra so often that
it will force you t9__l__3e<_;_qme proﬁcient with algebraicﬁtechniques. A11 equation may be simple in appearance, 4y = 15 (l) or rather complicated 2 = berx ﬂy (2) Solving a problem involving an algebraic equation often involves only
two steps:
Rearrange the equation to isolate the variable (x, y, '1', etc.) whose
value' is unknown. After this step, the equation will have the form Left side Right side unknown variable = X algebraic expression involving numbers
and variables whose values are known Mathematical equations can be readily rearranged by applying a
fundamental pprincl le of algebra which states that an equation
remains valid if the same operation is performed on both sides. I Add the same quantity to both sides. 
 Subtract the same quantity from both sides
' Multiply both sides by the same quantity 0 Divide both sides by the same quantity.  Raise both sides to the same power. 0 Extract the same mat of both sides. } "Golden rule " of algebra we preserve the equality.) , r'I The method to isolate a desired variable:
A. Remove terms and variables from one side of the equation by
performing the inverse operation (to undo the operation).
eg. to isolate x from (4/3 )x= 7 ,7
x is multiplied by 4, so we divide both sides by 4. :. 34c = 4
ab = (g7, : 7/“. .
q; x is divided by 3, so We midtipiyboﬂi sides by 3
B. When more than one term exists on the side with x,
transpose only entire terms. 5214132'7
C. When only a single term exists on the side containing x, perform
transpositions from the outermost parentheses and work inward.
D. If the unknown, )1, is in more than one term, ﬁrst attempt to place any terms containing 1: on one side of the equation, preferably in the numerator. I Then factor those terms to place it in only one location, if possible.
C2) Substitute numerical values for variables in the right— —hand side ofthe new equation. .
Using your calculator determine the value for the unknown variable For example consider equation (1) above. We can isolate the unknown, y,
on the left side by dividing both sides of the equation by the constant on the left, that is by4 “’3’ “ ‘5 5""
a 12
4 4
thus, .
' y = 15/4 \Mprefer badan 03%;?
or A» the durumthi I y=3.75 AuraL equivalent MW" 3 Next consider equation (2). Let us assume ﬁrst that the quantity whose value
»we wish to determine is b. To obtain it on the left side, switch sides of the equation:
be _‘ = ayz s'k...'. Then divide both sides by the terms other than b 011 the left, that' is, divide by ewﬂ' b = %
e
Ifwe are supplied with numerical values for a, y, c, and x, the value of b can
be computed =. 1a: 1+; :10 act—5' 'F“ 0' . u o . . . . .
.The 5&1 mvolvmg T is isolated by diwding both 81% by Q2: Subtracting 1 from both sides yields a term on the left which involves T only: As another example, suppose that "the unknown variable' in equation (2)is ' y rather than b. Dividing both sides of the original equation by. a isolates 312 on the left: beu
a y’= To obtain y, take the square root of both sides: + be'“
a edge brav
ml wer y Finally, consider a case in which the unknown variable in equation (2) is x.
We can isolate (on the left) the exponential term involving x by. switching sides of the equation _ 2
. be ex ay and then dividing both sides by b to obtain
e~cx__ _ ay’lb Since x occurs in an exponent, this equation must be manipulated further so.
that it takes the form x: ........ To accomplish this, take the logarithm to base e,that'1s, in, of both sides ofII
the equation.
Since ln(e""‘) —cx = in (ayZ/b) I = — we I
Therefore
Dividing by —c, we obtain our ﬁnal result I I
—In (0111/17) 6 “l: . bug“. X=c~135ucf As an examiple of these procedures, consider the equation
. 0.51.4 = Q1“ — 1/1“) Suppose that we must solve for A, given that X = 2, Q = 0.0048, and T = 4.8;
To obtain an equation for A, divide both sides by 0.5": Q2“ ' 117')
0.5’ Do the same thing to both sides
(if we treat both sides equally, Nada wilt. tmbulﬂar A: Now substitute the values for the variables on the right side: (0.0048110 —' 114.8) A _. 0.5z . Evaluating the two terms on top and the one on bottom gives (0.000023D)(0.792) '
0.25 A = 0.0011173 Next let us suppose that we must solve for T, given the values A = 4.20, Q =II
3.90, and x = 3. First switch sides of the equation to obtain T on the left: I Isaiah T: Q20 — 1/1) = use: Ag which yields
ll newer l tu‘
answer I l_W_05A 0.5‘A Q2
Rather than inverting the entire equation to obtain T, it is easier to evaluateI
the right side ﬁrst, otherwise a complicated equation_ will result —1 I
l .
I
I
I 1/1': . 0.53 x 4.20 I
'1” ‘W *
—1/T = —0.965
Inverting both sides gives
— = — 1/0965
=  1.04 Multiplying both sides by —1 eliminates the negative signs . 1..\
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 Fall '08
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