Beginning and Intermediate Algebra - Wallace Math Courses ...

Example 79.
A = πr 2 +πrs for s Solving fors, focus on what is added to the term with s
− πr 2 − πr 2 Subtractπr 2 from both sides
A − πr 2 =πrs s is multipied by πr
πr πr Divide both sides by πr
A −πr2 =s
πr
Our Solution
Again, we cannot reduce the πr in the numerator and denominator because of the
subtraction in the problem.
Formulas often have fractions in them and can be solved in much the same way
we solved with fractions before. First identify the LCD and then multiply each
term by the LCD. After we reduce there will be no more fractions in the problem
so we can solve like any general equation from there.
Example 80.
h = 2m
n
(n)h= (n)2m
n
for m To clear the fraction we use LCD =n
Multiply each term by n
nh =2m Reduce n with denominators
2 2 Divide both sides by 2
nh
=m
2
Our Solution
The same pattern can be seen when we have several fractions in our problem.
Example 81.
(b)a
b
a c
+ =e fora To clear the fraction we use LCD = b
b b
(b)c
+ =e(b) Multiply each term by b
b
a + c=eb Reduce b with denominators
− c − c Subtractcfrom both sides
a=eb − c Our Solution
Depending on the context of the problem we may find a formula that uses the
same letter, one capital, one lowercase. These represent different values and we
must be careful not to combine a capital variable with a lower case variable.
Example 82.
a = A
2 −b
forb Use LCD (2 −b) asagroup
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