88 Basic Engineering Mathematics
Cancelling gives
a
2
=
r p
(x + y)
Taking the square root of both sides gives
a =
rp
x + y
Whenever the letter required as the new subject
occurs more than once in the original formula, after
rearranging, factorizing will always be needed.
Problem 20. Make b the subject of the formula
a =
x − y
√
bd + be
Rearranging gives
x − y
√
bd + be
= a
Multiplying both sides by
√
bd + be gives
x − y = a
√
bd + be
or
a
√
bd + be = x − y
Dividing both sides by a gives
√
bd + be =
x − y
a
Squaring both sides gives
bd + be =
x − y
a
2
Factorizing the LHS gives
b(d + e) =
x − y
a
2
Dividing both sides by (d + e) gives
b =
x − y
a
2
(d + e)
or b =
(x − y) 2
a 2 (d + e)
Problem 21. If a =
b
1 + b
, make b the subject of
the formula
Rearranging gives
b
1 + b
= a
Multiplying both sides by (1 + b) gives
b = a(1 + b)
Removing the bracket gives
b = a + ab
Rearranging to obtain terms in b on the LHS gives
b − ab = a
Factorizing the LHS gives
b(1 − a) = a
Dividing both sides by (1 − a) gives b =
a
1 − a
Problem 22. Transpose the formula V =
Er
R + r
to make r the subject
Rearranging gives
Er
R + r
= V
Multiplying both sides by (R + r) gives
Er = V (R + r)
Removing the bracket gives
Er = V R + V r
Rearranging to obtain terms in r on the LHS gives
Er − V r = V R
Factorizing gives
r(E − V ) = V R
Dividing both sides by (E − V ) gives r =
V R
E − V
Problem 23. Transpose the formula
y =
pq 2
r + q 2 − t to make q the subject
Rearranging gives
pq 2
r + q 2 − t = y
and
pq 2
r + q 2 = y + t
Multiplying both sides by (r + q 2 ) gives
pq
2
= (r + q
2
)(y + t )
Removing brackets gives pq
2
= r y + rt + q
2 y + q
2 t
Rearranging to obtain terms in q on the LHS gives
pq
2
− q
2 y − q
2 t = r y + rt
Factorizing gives q
2
( p − y − t ) = r(y + t )
Dividing both sides by ( p − y − t ) gives
q
2
=
r(y + t )
( p − y − t )
Taking the square root of both sides gives
q =
r(y + t)
p − y − t
Problem 24. Given that
D
d
=
f + p
f − p
express p in terms of D, d and f