First-Degree Equations

A First-Degree Equation is the simplest type of integer numerical equation you can encounter.

The procedure for finding the solution of a first-degree equation consists of rewriting the equation in a familiar form and then checking the values of the coefficient of the unknown and the constant term.

In this lesson we will examine the solution procedure in detail and see which checks are required to determine whether an equation is determined.

Key Takeaways

A first-degree equation can be reduced to the form ax = b;
According to the values of $a$ and $b$ , the equation may be determined, indeterminate, or impossible.

First-Degree Equations

A first-degree equation is numerical (its coefficients are numbers) and polynomial, and it can be reduced to the following normal form:

a_1\,x + a_0 = 0

It is often convenient, using the first principle of equivalence, to rewrite it as

a\,x = b

where

a = a_1, \quad b = -a_0.

In other words, for reasons that will become clear in this lesson, it is useful to place the term containing the unknown on the left-hand side and the constant term on the right-hand side.

Definition

First-Degree Equations

A First-Degree Equation is an equation that is

Integer, i.e. the unknown does not appear in any denominator;
Numerical, i.e. all coefficients are numbers;

and can be reduced to the form

a\,x = b

The term $b$ is called the constant term.

Solving a first-degree equation

To solve a first-degree equation, the first step—using the principles of equivalence—is to reduce it to the form

a\,x = b.

Guidelines:

Move every term containing the unknown to the left-hand side;
Move every constant term to the right-hand side;
Perform the calculations so as to simplify and obtain the desired form.

Let us examine an example.

Example

Solve the following first-degree equation:

(2x - 3)(2x + 3) + 4(2 - x) = -(x - 9) + 4x(x - 2).

This equation is not yet in the desired form. First expand both sides so that each contains only algebraic sums of terms.

Notice that the first product is a notable product (sum times difference):

(2x - 3)(2x + 3) = 4x^{2} - 9.

Carrying out the calculations gives

\rightarrow\; 4x^{2} - 9 + 8 - 4x = -x + 9 + 4x^{2} - 8x.

Next, following the guidelines, bring all terms with the unknown to the left and all constants to the right. Before doing so, collect like terms:

\rightarrow\; 4x^{2} - 1 - 4x = -9x + 9 + 4x^{2}.

Now cancel identical terms on both sides using the cancellation rule:

\rightarrow\; \cancel{4x^{2}} - 1 - 4x = -9x + 9 + \cancel{4x^{2}}

\rightarrow\; -1 - 4x = -9x + 9.

Using the transport rule, move all terms with the unknown to the left and the constants to the right:

\rightarrow\; 9x - 4x = 1 + 9

\rightarrow\; 5x = 10.

The equation is now in the form ax = b.

After the equation has been rewritten in the form ax = b, one final step remains.

Solution of a first-degree equation

Once the equation is in the form ax = b, its solution depends on the values of $a$ and $b$ .

By applying the second principle of equivalence we may divide both sides by $a$ , provided $a\neq 0$ . The final solution is therefore

a\,x = b \quad\Longrightarrow\quad \frac{\cancel{a}}{\cancel{a}}x = \frac{b}{a} \quad\Longrightarrow\quad x = \frac{b}{a}.

In the previous example we obtained

5x = 10,

so a = 5 and b = 10. Because $a$ is non-zero we can apply the second principle and obtain

x = \frac{10}{5} \quad\Longrightarrow\quad x = 2.

We were able to apply the second principle because the coefficient $a$ is non-zero. This is a necessary condition for the equation to be determined.

Definition

Solution of a First-Degree Equation

A first-degree equation in the form ax = b is determined, i.e. admits a solution, if and only if the coefficient of the unknown is non-zero:

a \neq 0.

In this case the solution of a first-degree equation is

x = \frac{b}{a}.

Indeterminate and impossible first-degree equations

If the coefficient $a$ equals zero, we can no longer apply the second principle to solve the equation. The equation may then be indeterminate (infinitely many solutions) or impossible (no solution), depending on the value of the constant term $b$ .

First case: b = 0

If the constant term is zero the equation becomes

0\cdot x = 0.

In this case the equation admits infinitely many solutions: any value assigned to x satisfies the equality, since any number multiplied by zero equals zero. Thus the solution set is the set of real numbers and the equation is indeterminate:

Definition

Indeterminate First-Degree Equation

A first-degree equation in the form ax = b is indeterminate if

a = 0, \quad b = 0.

In this case the solution set is the entire set of real numbers:

S = \mathbb{R}.

Consider an example:

Example

Solve

x(x + 7) + 9 = x + (x + 3)^{2}.

First expand both sides:

\rightarrow\; x^{2} + 7x + 9 = x + x^{2} + 6x + 9.

Combine like terms:

\rightarrow\; x^{2} + 7x + 9 = x^{2} + 7x + 9.

Applying the cancellation rule eliminates the $x^2$ terms:

\rightarrow\; \cancel{x^{2}} + 7x + 9 = \cancel{x^{2}} + 7x + 9.

Moving terms yields

\rightarrow\; 7x - 7x = 9 - 9 \quad\Longrightarrow\quad 0x = 0.

Both $a$ and $b$ are zero, so the equation is indeterminate:

a = 0, \; b = 0 \quad\Longrightarrow\quad S = \mathbb{R}.

Second case: b ≠ 0

If the constant term is non-zero, the equation takes the form

0\cdot x = b.

Here the equation has no solution: no number multiplied by zero can yield a non-zero value. The equation is impossible:

Definition

Impossible First-Degree Equation

A first-degree equation in the form ax = b is impossible if

a = 0, \quad b \neq 0.

In this case the equation has no solution and the solution set is empty:

S = \varnothing.

Consider an example:

Example

Solve

8x - 3 + 2x = 6x + 1 + 4x.

Using the transport rule to gather terms:

\rightarrow\; 8x + 2x - 6x - 4x = 3 + 1 \quad\Longrightarrow\quad 0x = 4.

Here $a$ is zero while $b$ is non-zero, so the equation is impossible:

a = 0, \; b = 4 \quad\Longrightarrow\quad S = \varnothing.

Solution scheme

To summarise, solving a first-degree equation involves the following steps:

Hint

Using the principles of equivalence, rewrite the equation in the form:

$ax = b$
In order to do this, follow these steps:
- performing the necessary expansions and calculations;
- moving all terms with the unknown to the left-hand side;
- moving all constants to the right-hand side;
- simplifying by combining like terms.
Once the equation is in the desired form, examine the values of $a$ and $b$ :

$Solution of a first-degree equation$

In summary

In this lesson we have seen how to solve a first-degree equation. The procedure is

ax = b

followed by analysing the values of $a$ and $b$ to decide whether the equation is determined, indeterminate, or impossible:

If a ≠ 0, the solution is $x = \dfrac{b}{a}$ ;
If a = 0 and b = 0, the equation is indeterminate;
If a = 0 and b ≠ 0, the equation is impossible.