Literal First Degree Equation with One Variable

For an expression to be named as equation, it must have: equal sign, first and second member, and at least one variable. See the following examples, which are equations:

• 2x + 4 = 0
  2x + 4 → First member
  4 → Second member
  x → Variable

• 3y + 2 + 5y = y + 1
  3y + 2 + 5y → First member
  y + 1 → Second member
  y → Variable

One equation will be literal if it has all the characteristics described above and at least one letter that is not the variable, called a parameter and that takes on a numerical value. Some examples of literal equations are:

• 5ax + 10ax = 25
  5ax + 10ax → First member
  25 → Second member
  x → Variable
  a → Parameter

• 7aby + 11a = 5aby - 2
  7aby + 11a → First member
  5aby – 2 → Second member
  y → Variable
  a → Parameter
  b → Parameter

One literal equation will be of the first degree when the largest exponent the variable has is the number 1. Look:

• 2x + ax = 5 → 2x¹ + ax¹ = 5 → 1 is the degree of the literal equation with respect to variable x.

• 3aby + 5by = 2a → 3aby¹ + 5by¹ = 2a → 1 is the degree of the literal equation with respect to variable y.

To solve a literal equation of the first degree with one variable, we must isolate the term that represents the variable in one of the members of the equation so that, in the other member, we have its solution, which is represented by the parameter and some numerical value. Let's look at some literal equation resolutions:

Obtain the solution of the following literal equations:

The) ax + 2a = 2

B) 2by + 4 = 4b – 1

ç) 8c – 5cz = 2 + cz

Solution:

a) ax + 2a = 2

Variable: x
Parameter: a

ax + 2a = 2

ax = 2 - 2nd

x = 2 - 2nd
The

x = 2 - 2
The

x = 2nd^-1 – 2

First member (single variable): x
Second member and solution: 2nd^-1 – 2

b) 2by + 4 = 4b – 1

Variable: y
Parameter: b

5by + 4 = 5b - 1

5by = 5b - 1 - 4

5by = 5b - 5

y = 5b - 5
5b

y = 5b – 5
5b 5b

y = 1 - 1
B

y = 1 - 1b^{– 1}

First member (single variable): y
Second member and solution: 1 – 1b^{– 1}
c) 8ac – 5acz = 2 + cz

Variable: z
Parameters: a, c

8c – 5acz = 2 + acz

- 5acz – acz = 2 – 8c

- 6 acz = 2 - 8c

- z = 2 - 8c. (- 1)
6ac

- (- z) = - (2 - 8c)
6ac

+ z = - 2 + 8 c
6ac

First member (single variable): z
Second member and solution: - 2 + 8 c
6ac

By Naysa Oliveira
Graduated in Mathematics

Source: Brazil School - https://brasilescola.uol.com.br/matematica/equacao-literal-primeiro-grau-com-uma-variavel.htm