Literal Equations
A literal equation is an equation that contains at least one letter in addition to the unknown variable or variables.
The letter represents an unknown constant or parameter that affects the equation's behavior.
Here's a simple example of a literal equation:
$$ kx + 2 = 5 $$
In this expression, k is a parameter, while x is the unknown variable.
What's the difference between parameters and unknowns? Parameters are fixed values determined externally; they define the general form of the equation. Unknowns, on the other hand, are the quantities we solve for within the equation itself.
Literal equations can be either whole (without fractions) or fractional.
They are widely used in mathematics and physics to describe relationships between variables in a general, symbolic form.
How to Solve a Literal Equation
Solving a literal equation requires a careful discussion of the possible values that can be assigned to the parameters - that is, to the letters representing constants.
Note: For fractional literal equations, it's also essential to consider the domain of definition - the values for which the algebraic fractions are actually defined.
A Practical Example
Consider the following whole literal equation:
$$ kx + k = 0 $$
We can rewrite it in the standard form Ax = B.
In other words, we factor out the variable on the left-hand side and move the constant term to the right-hand side:
$$ kx = -k $$
Discussion:
Let's now analyze the possible values of the parameter k.
Case 1: k = 0 → the equation is indeterminate
$$ kx = -k $$
$$ 0x = 0 $$
When k = 0, the equation becomes 0 = 0, which is true for any x, so it's indeterminate.
Case 2: k ≠ 0 → the equation is determinate
$$ kx = -k $$
We isolate the variable:
$$ \frac{kx}{k} = \frac{-k}{k} $$
$$ \require{cancel} \frac{\cancel{k}x}{\cancel{k}} = \frac{-\cancel{k}}{\cancel{k}} $$
$$ x = -1 $$
Therefore, when k ≠ 0, the equation has a unique solution: x = - 1.
Example 2
Now consider a fractional literal equation:
$$ \frac{x-1}{k} + \frac{2x+3}{4k} = \frac{x}{4} $$
Since this is a fractional equation, we first need to determine its domain of existence.
The denominators become zero when k = 0, so we must exclude that value.
Thus, the condition of existence (C.E.) is k ≠ 0.
$$ C.E. \ k \ne 0 $$
Next, we collect the variable x and rewrite the equation in the form Ax = B:
$$ \frac{x-1}{k} + \frac{2x+3}{4k} = \frac{x}{4} $$
$$ \frac{2x+3+4(x-1)}{4k} - \frac{x}{4} = 0 $$
$$ \frac{2x+3+4(x-1)-kx}{4k} = 0 $$
$$ \frac{2x+3+4x-4-kx}{4k} = 0 $$
$$ \frac{6x-1-kx}{4k} = 0 $$
We now eliminate the denominator by applying the principles of equivalence - multiplying both sides by 4k:
$$ \frac{6x-1-kx}{4k} \cdot 4k = 0 \cdot 4k $$
$$ \frac{6x-1-kx}{\cancel{4k}} \cdot \cancel{4k} = 0 $$
$$ 6x - 1 - kx = 0 $$
$$ 6x - kx = 1 $$
$$ x(6 - k) = 1 $$
Discussion:
Let's discuss the possible values of k.
Case 1: k = 6 → the equation is impossible
$$ x(6 - k) = 1 $$
$$ x(6 - 6) = 1 $$
$$ 0 = 1 $$
When k = 6, the equation has no solution, so it's impossible.
Case 2: k = 0 → violates the initial condition
$$ x(6 - k) = 1 $$
$$ C.E. \ k \ne 0 $$
When k = 0, the equation is undefined because it doesn't satisfy the condition of existence.
Case 3: k ≠ 6 and k ≠ 0 → the equation is determinate
$$ x(6 - k) = 1 $$
We isolate x:
$$ \frac{x(6 - k)}{6 - k} = \frac{1}{6 - k} $$
$$ \frac{x \cancel{(6 - k)}}{\cancel{6 - k}} = \frac{1}{6 - k} $$
$$ x = \frac{1}{6 - k} $$
In this case, the equation is determinate, with a single solution: x = 1 / (6 - k).
And so on.
