 Example 1:
Solve the formula
P = 2L + 2W
for W.
Solution to Example 1

Given
P = 2L + 2W

we first isolate the term containing W : add 2L to both sides of the equation
P  2L = 2L + 2W  2L

Simplify to obtain
P  2L = 2W

Divide both sides by 2 to obtain W.
W = (P 2L) / 2
Example 2:
Solve the formula
H = √ ( x^{ 2} + y^{ 2})
for y, where H, x and y are a positive real numbers and H is greater than x and greater than y.
Solution to Example 2

Given
H = √ ( x^{ 2} + y^{ 2})

Square both sides
H^{ 2} = x^{ 2} + y^{ 2}

Add  x^{ 2} to both sides and simplify
H^{ 2}  x^{ 2} = x^{ 2} + y^{ 2}  x^{ 2}
H^{ 2}  x^{ 2} = y^{ 2}

Solve for y taking the square root
y = ~+mn~ √ (H^{ 2}  x^{ 2})

Since y is a positive real number, then y is given by
y = + √ (H^{ 2}  x^{ 2})
Example 3:
Express F in terms of C in the formula
C = (5 / 9)(F  32)
.
Solution to Example 3
C = (5 / 9)(F  32)

Multiply both sides of the formula by 9 / 5
(9 / 5) C = (9 / 5)(5 / 9)(F  32)

and simplify
(9 / 5) C = (F  32)

Add 32 to both sides of the formula.
(9 / 5) C + 32 = F

The formula F = (9 / 5) C + 32 expresses F in terms of C.
Example 4:
Express y in terms of x in the equation
a x + b y = c , with b not equal to zero.
.
Solution to Example 4
a x + b y = c

Add  a x to both sides of the equation
a x + b y  a x = c  a x
b y =  a x + c

Divide both sides by b.
y =  (a / b) x + c / b
Exercises:
Solve each of the formulas below for the indicated variable.(see answers below).
 A = W L , for L.
 y = m x + b , for x.
 A = (1 / 2)(B + a) , for a.
 S = 2 π r h , for r.
 F = (9 / 5)C + 32 , for C.
 1 / x = 1 / y + 1 / z , for y.
Answers to Above Exercises:
Solve each of the formulas below for the indicated variable.
 L = A / W
 x = (y  b) / m , for m not equal to zero.
 a = 2 A  B
 r = S / (2 π h)
 C = (5 / 9)(F  32)
 y = (x z) / (z  x) , for z not equal to x.
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