Math Lesson 9.1.5 - First Order Equations with One Variable

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Welcome to our Math lesson on First Order Equations with One Variable, this is the fifth lesson of our suite of math lessons covering the topic of Variables, Coefficients and Constants. First Order Equations with One Variable, you can find links to the other lessons within this tutorial and access additional Math learning resources below this lesson.

First Order Equations with One Variable

The simplest type of equation is the first - order equation with one variable. As the name implies, such equations have a single variable, which is in the first power. The terms 'first order' or 'first degree' are used to describe equations (or polynomials as we will see in chapter 11) where variables appear in the first power only. For example, 2x - 4 = 0 is a first - order equation with one degree, as there is a single variable (x) in the equation and this variable is in the first power (x1 = x). On the other hand, equations such as 5x + 2y = 6 and x² + 3 = 7 are not of the first order with one variable as the first equation contains two variables (x and y) while the second equation is not of the first order as its variable (x) is in raised to the second power (it is a second - order equation).

Example 5

Are the following equations first - order equations with one variable? If yes, write them in the regular form.

1. 3 - 2x = 5x - 1
2. x · 3x + 2x = 11
3. y - 4x = 3x - 2 - 7x
4. a - 3b = a + 4b - 1

Solution 5

From theory, it is a known fact that a first - order equation must contain a single variable in the first power. However, to know whether the given belong to this type we must first complete some operations to turn them into the simplest form. Then, we must write it in the form E(x) = 0.

1. We have
   3 - 2x = 5x - 1
   3 - 2x - 5x + 1 = 0
   - 7x + 4 = 0
   It is obvious that this equation is a first - order type with one variable.
2. We have
   x ∙ 3x + 2x = 11
   3x² + 2x - 11 = 0
   This is an equation with one variable but it is not of the first order, as the highest power of the variable is 2. This is a second - order equation with one variable, otherwise known as a quadratic equation, briefly discussed in tutorial 6.4. We will deal more extensively with such equations in the next tutorials of the current chapter.
3. At first glance it seems to be a first - order equation with two variables, x and y. However, after doing the operations with like terms and the simplifications resulting from these operations yields
   y - 4x = 3x - 2 - 7x
   y - 4x = - 4x - 2
   y - 4x + 4x + 2 = 0
   y + 2 = 0
   This is a first - order equation with one variable (y).
4. Again, it seems like a first - order equation with two variables (a and b), but we will see during the operations that this is not true. We have
   a - 3b = a + 4b - 1
   a - 3b - a - 4b + 1 = 0
   - 7b + 1 = 0
   Hence, given that the only variable in this equation is b and it is in the first power, this is an example of first - order equations with one variable.

Solving First Order Equations with One Variable

Solving first - order equations with one variable means isolating the variable and expressing it in terms of the other elements of the equation (numbers and non - variable letters).

The letter that represents more frequently the variable in equations with one variable is 'x'. Therefore, the general formula of the first - order equation with one variable is

where a and b are numbers (a is a coefficient and b is a constant).

Solving a first - order equation with one variable means calculating the value of its variable. In math language, this process is called "finding the root of the equation". This has nothing to do with the concept of roots explained in the seventh chapter.

The necessary thing to do when solving such equations is to isolate the variable. Thus, we have

Dividing both sides by a yields

For example, the solution of the equation

is 2 because

Example 6

Solve the following first - order equations with one variable:

1. 2 - 5x = 17
2. 3x - 6 = 4y - 5 + 3x

Solution 6

1. First, we arrange the equation to show it in the form ax + b = 0. Thus,
   2 - 5x = 17
   2 - 5x - 17 = 0
   - 5x - 15 = 0
   We have a = -5 and b = -15. Thus,
   x = - b/a
   = - -15/-5
   = - 3
2. Again, first we arrange the equation to show it in the form ax + b = 0. Thus,
   3x - 6 = 4y - 5 + 3x
   3x - 3x - 4y + 5 = 0
   - 4y + 5 = 0
   We have a = -4 and b = 5. Thus,
   x = - b/a
   = -5/-4
   = 5/4

Transformations that can be made in First - Order Equations with One Variable

When solving a first - order equation with one variable, the following transformations can take place:

1. If we add or subtract the same number, variable or term from both sides of an equation, we obtain an equivalent equation.
   For example,
   2x - 3 = 4; 2x + 3x - 3 = 4 + 3x; 2x - 3 - 1 = 4 - 1; etc.
   are all equivalent equations.
2. Multiplying or dividing both sides of an equation by the same non - zero number gives an equivalent equation.
   For example,
   1 - 3x = 6; 1 - 3x/6 = 6/6; 2 ∙ (1 - 3x) = 2 ∙ 6; etc.
   are all equivalent equations.

Such transformations help us understand why we change the sign of a number, variable or term when sending it on the other side of the equation. Let's clarify this point through a couple of examples.

Example 7

Solve the equations below by explaining each step.

1. 2 - 5x = x/3 + 1
2. 5/2x + 3 = 4/x

for x ≠ 0.

Solution 7

1. We have
   2 - 5x = x/3 + 1
   Multiply both sides by 3 to eliminate the fraction.
   3 ∙ (2 - 5x) = 3 ∙ (x/3 + 1)
   Expand and then make operations with like terms.
   3 ∙ 2 + 3 ∙ ( - 5x) = 3 ∙ x/3 + 3 ∙ 1
   6 - 15x = x + 3
   Subtract 3 from both sides to eliminate numbers from the right side.
   6 - 15x - 3 = x + 3 - 3
   3 - 15x = x
   Subtract x from both sides to eliminate variables from the right side.
   3 - 15x - x = x - x
   3 - 16x = 0
   Now, write the equation in the form ax + b = 0.
   - 16x + 3 = 0
   We have a = -16 and b = 3. Thus, the value of the variable x is
   x = - b/a
   = - -16/3
   = 16/3
2. We have
   5/2x + 3 = 4/x
   Multiply both sides by 2x to eliminate fractions.
   2x ∙ (5/2x + 3) = 2x ∙ 4/x
   Expand, simplify and then make operations with like terms.
   2x ∙ 5/2x + 2x ∙ 3 = 2x ∙ 4/x
   5 + 6x = 2 ∙ 4
   5 + 6x = 8
   Subtract 8 from both sides to eliminate numbers from the right side.
   5 + 6x - 8 = 8 - 8
   6x - 3 = 0
   We have a = 6 and b = -3. Thus, the value of the variable x is
   x = - b/a
   = - -3/6
   = 1/2

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