The process of solving literal equations is very similar to solving linear equations. Our goal is to isolate one variable on one side of the equation while the rest (which may include other variables and constants) are on the opposite side.

**Example 1:** Solve the literal equation below for \(b\).

$${{a - b} \over b} = 1$$

**STEP 1:** Multipy both sides by \(b\).

$$a - b = b$$

**STEP 2:** Add \(b\) on both sides of the equation.

$$\eqalign{

a - b + b &= b + b \cr

a &= 2b \cr} $$

**STEP 3:** To solve for \(b\), we divide both sides by 2.

$$\eqalign{

{a \over 2} &= {{2b} \over 2} \cr

{a \over 2} &= b \cr} $$

Therefore,

$$b = {a \over 2}$$

**Example 2:** Solve the literal equation below for \(x\).

$${{x + 3} \over {4y - 8}} = {x \over 2}$$

**STEP 1:** Perform cross multiplication.

$$2\left( {x + 3} \right) = x\left( {4y - 8} \right)$$

**STEP 2:** Distribute the term outside the parenthesis to the terms inside.

$$2x + 6 = 4xy - 8x$$

**STEP 3:** Move all the x's to the right side by subtracting 2x on both sides.

$$2x - 2x + 6 = 4xy - 8x - 2x$$

$$6 = 4xy - 10x$$

**STEP 4:** Factor out x on the right side,

$$6 = x\left( {4y - 10} \right)$$

**STEP 5:** Divide both sides by \(4y-10\).

$${6 \over {4y - 10}} = x$$

**STEP 6:** Cancel out the common factor of 2 between the numerator and denominator.

$$\eqalign{

{{2\left( 3 \right)} \over {2\left( {2y - 5} \right)}} &= x \cr

{3 \over {2y - 5}} &= x \cr} $$