First degree equations formula, how to solve them, example, exercises

The first degree or linear equations with an unknown are those that can be expressed as the sum of two terms, as follows:

ax + b = 0

Where a and b, with to ≠ 0, are real numbers R or also complex C. To solve it, terms are transposed, which means changing terms from one side of the equality to the other.

To solve the unknown, the term + b is transposed, which must go to the right side of the equality with changed sign.

ax = -b

Then the value of x is cleared, like this:

x = - b / a

As an example we are going to solve the following equation:

6x - 5 = 4

We transpose the -5 term to the right side with a changed sign:

6x = 4 + 5

This is equivalent to adding 5 to both sides of the original equation:

6x - 5 + 5 = 4 + 5 → 6x = 9

And now we solve the unknown "x":

x = 9/6 = 3/2

Which is equivalent to dividing both sides of the equality by 6. So we can use the following to obtain the solution:

-You can add or subtract the same quantity from both sides of the equality in an equation, without altering it.

-You can also multiply (or divide) by the same amount all the terms both to the left and to the right of the equation.

-And if both members of an equation are raised to the same power, the equality is not altered either.

Article index

• 1 How to solve first degree equations
  • 1.1 Graphic interpretation
• 2 Examples of simple linear equations   
  • 2.1 Integer equations
  • 2.2 Fractional equations
  • 2.3 Literal equations
  • 2.4 Systems of first degree equations 
  • 2.5 Linear equations with absolute value
• 3 Simple solved exercises
  • 3.1 - Exercise 1
  • 3.2 - Exercise 2
  • 3.3 - Exercise 3
• 4 References

How to solve first degree equations

The solution of an equation of the first degree is also known as its root. It is the value of x that converts the original expression to an equality. For example in:

5x = 8x - 15

If we substitute x = 5 in this equation, we get:

5⋅5 = 8⋅5 - 15

25 = 40 - 15

25 = 25

Since linear equations of the first degree come in many forms, which are sometimes not obvious, there are a series of general rules that include various algebraic manipulations, in order to find the value of the unknown:

-First of all, if there are indicated operations, they should be carried out.

-Grouping symbols such as parentheses, square brackets, and braces, if any, should be removed keeping the appropriate signs.

-The terms are transposed to place all those that contain the unknown on one side of the equality, and those that do not contain it on the other.

-Then all like terms are reduced to arrive at the form ax = -b.

-And the last step is to clear the unknown.

Graphic interpretation

The first degree equation raised at the beginning can be derived from the equation of the line y = mx + c, making y = 0. The resulting value of x corresponds to the intersection of the line with the horizontal axis.

In the following figure there are three lines. Starting with the green line, whose equation is:

y = 2x - 6

Making y = 0 in the equation of the line the first degree equation is obtained:

2x - 6 = 0

Whose solution is x = 6/2 = 3. Now when we detail the graph, it is easy to realize that in effect, the line intersects the horizontal axis at x = 3.

The blue line intersects the x-axis at x = 5, which is the solution to the equation -x + 5 = 0. Finally, the line whose equation is y = 0.5x + 2 intersects the x-axis at x = - 4, which is easily seen from the equation of the first degree:

0.5 x + 2 = 0

x = 2 / 0.5 = 4

Examples of simple linear equations   

Integer equations

They are those in whose terms there are no denominators, for example:

21 - 6x = 27 - 8x

Your solution is:

-6x + 8x = 27 - 21

2x = 6

x = 3

Fractional equations

These equations contain at least one denominator other than 1. To solve them, it is advisable to multiply all the terms by the least common multiple (LCM) of the denominators, in order to eliminate them.

The following equation is fractional type:

Since these numbers are small, it is not difficult to see that m.c.m (6, 8,12) = 24. This result is easily obtained by expressing the numbers as a product of prime numbers or their powers, let's see:

6 = 3.2

8 = 2³

12 = 2^two⋅3

The least common multiple is determined by multiplying the common and uncommon factors of 6, 8, and 12 with their greatest exponent, then:

lcm (6,8,12) = 2³ ⋅3 = 8 × 3 = 24

Since we have the least common multiple, we must multiply it by each of the terms of the equation:

4 (x + 5) -3 (2x + 3) = 2 (1-5x)

We make use of the distributive property:

4x + 20 - 6x -9 = 2 - 10x

All the terms that contain the unknown "x" are grouped on the left side of the equality, leaving the independent or numerical terms on the right side:

4x - 6x + 10 x = 2 +9 - 20

8x = -9

x = - 9/8

Literal equations

They are linear equations with one unknown, which however are accompanied by literal coefficients (letters). These letters are treated in the same way as numbers. An example of a literal first degree equation is:

-3ax + 2a = 5x - b

This equation is solved in the same way as if the independent terms and coefficients were numerical:

-3ax - 5x = - b - 2a

Factoring the unknown "x":

x (-3a - 5) = - b - 2a

x = (- b - 2a) / (-3a - 5) → x = (2a + b) / (3a + 5)

Systems of equations of the first degree 

Systems of equations consist of a set of equations with two or more unknowns. The solution of the system consists of values ​​that satisfy the equations simultaneously and to determine it unequivocally, there must be an equation for each unknown.

The general form of a system of m linear equations with n unknowns is:

to_elevenx₁ + to₁₂x_two +… to_1nx_n = b₁
to_twenty-onex₁ + to₂₂x_two +… to_2nx_n = b_two
...
to_m1x₁ + to_m2x_two +… to_mnx_n = b_m

If the system has a solution, it is said to be compatible determined, when there is an infinite set of values ​​that satisfy it is indeterminate compatible, and finally if it has no solution then it is incompatible.

In solving systems of linear equations, several methods are used: reduction, substitution, equalization, graphical methods, Gauss-Jordan elimination and the use of determinants are among the most used. But there are other algorithms to reach the solution, more convenient for systems with many equations and unknowns.

An example of a system of linear equations with two unknowns is:

8x - 5 = 7y - 9
6x = 3y + 6

The solution of this system is presented later in the solved exercises section..

Linear equations with absolute value

The absolute value of a real number is the distance between its location on the number line and 0 on the number line. As it is a distance, its value is always positive.

The absolute value of a number is denoted by the modulo bars: │x│. The absolute value of a positive or negative number is always positive, for example:

│ + 8│ = 8

│-3│ = 3

In an absolute value equation, the unknown is between modulus bars. Let's consider the following simple equation:

│x│ = 10

There are two possibilities, the first is that x is a positive number, in which case we have:

x = 10

And the other possibility is that x is a negative number, in this case:

x = -10

These are the solutions of this equation. Now let's look at a different example:

│x + 6│ = 11

The amount inside the bars can be positive, so:

x + 6 = 11

x = 11 -6 = 5

Or it can be negative. In that case:

-(x + 6) = 11

-x - 6 = 11 ⇒ -x = 11 + 6 = 17

And the value of the unknown is:

x = -17

This absolute value equation therefore has two solutions: x₁ = 5 and x_two = -17. We can check that both solutions lead to an equality in the original equation:

│5 + 6│ = 11

│11│ = 11

Y

│-17 + 6│ = 11

│-11│ = 11

Simple solved exercises

- Exercise 1

Solve the following system of linear equations with two unknowns:

8x - 5 = 7y -9
6x = 3y + 6

Solution

As it is proposed, this system is ideal for using the substitution method, since in the second equation the unknown x is almost ready for clearance:

x = (3y + 6) / 6

And it can be immediately substituted into the first equation, which then becomes a first degree equation with unknown "y":

8 [(3y + 6) / 6] - 5 = 7y - 9

The denominator can be removed by multiplying each term by 6:

6. 8⋅ [(3y + 6) / 6] - 6.5 = 6 .7y- 6. 9

8⋅ (3y + 6) - 30 = 42y - 54

Applying the distributive property in the first term to the right of the equality:

24y + 48 -30 = 42y - 54 ⇒ 24y + 18 = 42y - 54

The equation can be simplified, since all coefficients are multiples of 6:

4y + 3 = 7y - 9

-3y = -12

y = 4

With this result we go to the clearance of x:

x = (3y +6) / 6 → x = (12 + 6) / 6 = 3

- Exercise 2

Solve the following equation:

Solution

Products appear in this equation, and following the instructions given at the beginning, they must be developed first:

3x - 10x +14 = 5x + 36x + 12

Then all the terms that contain the unknowns are taken to the left side of the equality, and to the right side will be the independent terms:

3x - 10x - 5x - 36x = 12 - 14

-48x = -2

x = 1/24

- Exercise 3

Adding the three interior angles of a triangle gives 180º. The major exceeds the minor by 35º, and this in turn exceeds by 20º the difference between the major and the median. What are the angles?

Solution

We will call “x” to the greater angle, “y” to the middle one and “z” to the smallest. When the statement affirms that the sum of them is 180º, it can be written:

x + y + z = 180

Then we know that the greater exceeds the lesser by 35º, we can write this like this:

x = z + 35

Finally, the smallest exceeds the difference between the largest and the medium by 20º:

z = x - y + 20

We have a system of 3 equations and 3 unknowns:

x + y + z = 180

x = z + 35

z = x - y + 20

Solving for z from the first equation we have:

z = 180 - x - y

Matching with the third:

180 - x - y = x - y + 20

Passing the unknowns to the left side as always:

-x - y - x + y = 20 - 180

The "y" is canceled and remains:

-2x = - 160

x = 80º

From the second equation we find the value of z:

z = x - 35 = 80 - 35 = 45º

And the value of y is found from the first or third:

y = 180 - x - z = 180 - 80 - 45 = 55º

References

1. Baldor. 1977. Elementary Algebra. Venezuelan Cultural Editions.
2. Monterey Institute. Equations, inequalities and absolute value. Recovered from: montereyinstitute.org.
3. Online teacher. Classification of linear or first degree equations. Recovered from: profesorenlinea.cl.
4. Hoffman, J. Selection of Mathematics Topics. Volume 2.
5. Jiménez, R. 2008. Algebra. Prentice hall.
6. Zill, D. 1984. Algebra and Trigonometry. Mcgraw hill.