SOLVING LITERAL EQUATIONS

What is a literal equation?

A literal equation is an equation that consists primarily of letters.

Formulas are an example of literal equations.

Each variable in the equation "literally" represents an important part of the whole relationship expressed by the equation.

How to solve a literal equations ?

Step 1 :

Identify the variable that we need to isolate.

Step 2 :

Using inverse operations, we have remove the remaining terms all around the variable that we have targeted.

Step 3 :

We apply all the rules of algebra while solving equations.

Solve A = (a + b)/2 for b

Solve the following :

Example 1 :

Solve V = lwh for w

Solution :

Given,

V = lwh

Solving for w.

Divide by lh on both sides,

V/lh = w

w = V/lh

So, the solution of w is V/lh.

Example 2 :

Solve m = (y₂ - y₁)/(x₂ - x₁) for y₂

Solution :

Given,

m = (y₂ - y₁)/(x₂ - x₁)

Solving for y₂.

Multiply by (x₂ - x₁) on both sides,

m(x₂ - x₁) = y₂ - y₁

mx₂ - mx₁ = y₂ - y₁

Add y₁ on both sides,

mx₂ - mx₁ + y₁ = y₂

y₂ = mx₂ - mx₁ + y₁

So, the solution of y₂ is mx₂ - mx₁ + y_1.

Example 3 :

Solve ax + by = c for y

Solution :

Given,

 ax + by = c

Solving for y.

Subtract ax on both sides,

by = c - ax

Divide by b on both sides,

y = (c - ax)/b

So, the solution of y is (c - ax)/b.

Example 4 :

Solve A = (a + b + c + d)/4 for c

Solution :

Given,

A = (a + b + c + d)/4

Solving for c.

Multiply by 4 on both sides,

4A = a + b + c + d

Along with c, a, b and d are added. So, subtract a, b and d on both sides.

4A - a - b - d = c

c = 4A - a - b - d

So, the solution of c is 4A - a - b - d.

Example 5 :

Solve S = 2(lw + lh + wh) for w.

Solution :

Given,

S = 2(lw + lh + wh)

S = 2lw + 2lh + 2wh

By combining like terms,

S = 2lw + 2wh + 2lh

S = 2w(l + h) + 2lh

Solving for w.

Subtract by 2lh on both sides,

S - 2lh = 2w(l + h)

Divide by 2(l + h) both sides,

(S - 2lh)/[2(l + h)] = w

w = (S - 2lh)/(2l + 2h)

So, the solution of c is (S - 2lh)/(2l + 2h).

Example 6 :

Solve P = 2(l + w) for l

Solution :

Given,

P = 2(l + w)

P = 2l + 2w

Solving for l.

Subtract 2w on both sides,

P - 2w = 2l

Divide by 2 on both sides,

(P - 2w)/2 = l

So, the solution of l is (P - 2w)/2.

Example 7 :

Solve d = c/π for π

Solution :

Given,

d = c/π

πd = c

Solving for π.

Divide by d on both sides,

π = c/d

Example 8 :

Solve 5t - 2r = 25 for t

Solution :

Given,

5t - 2r = 25

Solving for t.

Add 2r on both sides, 

5t = 25 + 2r

Divide by 5 on both sides, 

t = 25/5 + 2/5r

t = 5 + 2/5r

Example 9 :

Solve S = R - rR for R.

Solution :

Given,

S = R - rR

S = R(1 - r)

Solving for R.

Divide by (1 - r) on both sides,

S/(1 - r) = R

Example 10 :

Solve V = 1/3πh²(3r - h) for r

Solution :

Given,

V = 1/3πh²(3r - h)

By using the distributive property,

V = 3/3πh²r - 1/3πh³

V = πh²r - πh³/3

By taking the least common multiple,

V = (3πh²r - πh³)/3

Solving for r.

Multiply by 3 on both sides,

3V = 3πh²r - πh³

Add πh³ on both sides,

3V + πh³ = 3πh²r

Divide by 3πh² on both sides,

(3V + πh³)/3πh² = r

Example 11 :

Solve A = 1/2nal for n

Solution :

Given,

 A = 1/2nal

Solving for n.

Multiply by 2 on both sides,

2A = nal

Divide by al on both sides,

2A/al = n

Example 12 :

Solve (p₁v₁)/T₁ = (p₂v₂)/T₂ for T₁

Solution :

Given,

(p₁v₁)/T₁ = (p₂v₂)/T₂

Solving for T_1.

Multiply by T₁ on both sides,

p₁v₁ = T₁(p₂v₂)/T₂

Multiply by T₂ on both sides,

T₂(p₁v₁) = T₁(p₂v₂)

Divide by p₂v₂ on both sides,

T₂(p₁v₁)/(p₂v₂) = T₁

Example 13 :

Solve F = (gm₁m₂)/d² for g

Solution :

Given,

F = (gm₁m₂)/d²

Solving for g.

Multiply by d² on both sides,

Fd² = gm₁m₂

Divide by m₁m₂ on both sides,

Fd²/m₁m₂ = g

Example 14 :

Solve (12ds)/w = CD for w

Solution :

Given,

(12ds)/w = CD

Solving for w.

Multiply by w on both sides,

12ds = wCD

Divide by CD on both sides,

12ds/CD = w

Example 15 :

Solve A = 1/2bh for b

Solution :

Given,

A = 1/2bh

Solving for b.

Multiply by 2 on both sides,

2A = bh

Divide by h on both sides,

2A/h = b