SOLVING LITERAL EQUATIONS

A literal equation is, simply put, an equation that has a lot of letters or variables.

For example,

A = π“ β‹… w 

(The formula for finding the area of a rectangle)

E = mc2

(Einstein’s Theory of Relativity)

are both literal equations. 

When a literal equation is given, we would often be asked to solve the equation for a given variable. The goal is to isolate that given variable. The process is the same process that we use to solve linear equations; the only difference is that we will be working with a lot more letters, and we may not be able to simplify as much as we can with linear equations. 

Solved Examples

Example 1 :

Solve for w in the formula for area of a rectangle : 

A = π“ β‹… w

Solution : 

A = π“ β‹… w

Divide each side by 𝓁.

ᴬ⁄𝓁 = w

Example 2 : 

Solve for c in the formula for Einstein’s Theory of Relativity : 

E = mc2

Solution : 

E = mc2

Divide each side by m.

ᴱ⁄m = c2

Take square root on each side. 

√(ᴱ⁄m) = √c2

√(ᴱ⁄m) = c

Example 3 : 

Solve for h in the formula for the surface area of a right cylinder :

S = 2Ο€r(h + r)

Solution : 

S = 2Ο€r(h + r)

Use distributive property of multiplication over addition. 

S = 2Ο€rh + 2Ο€r2

Subtract 2Ο€r2 from each side. 

S - 2Ο€r= 2Ο€rh

Divide each side by 2Ο€r. 

Example 4 : 

Solve for r in the formula for volume of sphere :

V = ⁴⁄₃ β‹… Ο€r3

Solution : 

V = β΄β„₃ β‹… Ο€r3

Multiply each side by 3/4.

³ⱽ⁄₄ = Ο€r3

Divide each side by Ο€.

³ⱽ⁄₄π = r3

Take cube root on each side.

3√(³ⱽ⁄₄π) = 3√r3

3√(³ⱽ⁄₄π) = r

Example 5 :

Solve for w in the formula for perimeter of the rectangle :

P = 2(𝓁 + w)

Solution :

P = 2(𝓁 + w)

Use distributive property of multiplication over addition.

P = 2𝓁 + 2w

Subtract 2𝓁 from each side.

P - 2𝓁 = 2w

Divide each side by 2.

⁽ᴾ ⁻ ²𝓁⁾⁄₂ = w

Example 6 : 

Solve for a :

Q = 3a + 5ac

Solution : 

Q = 3a + 5ac

Factor a out from (3a + 5ac). 

Q = a(3 + 5c)

Divide each side by (3 + 5c). 

Q⁄₍₃ β‚Š β‚…cβ‚Ž = a

Example 7 : 

Solve for b1 :

A = Β½ β‹… h(b1 + b2)

Solution : 

A = Β½ β‹… h(b1 + b2)

A = Κ°β„β‚‚ β‹… (b1 + b2)

Multiply each side by ²⁄h.

²ᴬ⁄h = b1 + b2

Subtract b2 from each side. 

²ᴬ⁄h - bb1

Example 8 : 

Solve for b :

a2 + b2 = c2

Solution : 

a2 + b2 = c2

Subtract a2 from each side. 

b2 = c- a2

Take square root on each side. 

√b2 = √(c- a2)

b = √(c- a2)

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