What is an equation with one unknown?

Equations are algebraic expressions containing numbers and unknowns. It is important to differentiate these two groups: the numbers are fixed values while the unknowns, as their name indicates, represent unknown values (at least at the beginning), and in most cases we are asked to find out what this value is.
For example:

What do we do with the equations?

When we are given an exercise that contains an equation with an unknown, our goal is to solve the equation, that is, to find a solution to the equation. What does it mean to find the solution to an equation? The idea is to find the value of the unknown with the goal of making both sides of the equation equal.

When we have equations that have the same solution, they will be called equivalent equations.

When first degree equations include fractions, and the unknown is in the denominator, it is important to keep in mind the domain of the function

examples with solutions for first-degree equations

Exercise #1

Solve the equation

$20:4x=5$

Step-by-Step Solution

To solve the exercise, we first rewrite the entire division as a fraction:

$\frac{20}{4x}=5$

Actually, we didn't have to do this step, but it's more convenient for the rest of the process.

To get rid of the fraction, we multiply both sides of the equation by the denominator, 4X.

20=5*4X

20=20X

Now we can reduce both sides of the equation by 20 and we will arrive at the result of:

X=1

$x=1$

Exercise #2

Solve the equation

$5x-15=30$

Step-by-Step Solution

We start by moving the sections:

5X-15 = 30
5X = 30+15

5X = 45

Now we divide by 5

X = 9

$x=9$

Exercise #3

Find the value of the parameter X

$\frac{1}{3}x+\frac{5}{6}=-\frac{1}{6}$

Step-by-Step Solution

First, we will arrange the equation so that we have variables on one side and numbers on the other side.

Therefore, we will move $\frac{5}{6}$ to the other side, and we will get

$\frac{1}{3}x=-\frac{1}{6}-\frac{5}{6}$

Note that the two fractions on the right side share the same denominator, so you can subtract them:

$\frac{1}{3}x=-\frac{6}{6}$

Observe the minus sign on the right side!

$\frac{1}{3}x=-1$

Now, we will try to get rid of the denominator, we will do this by multiplying the entire exercise by the denominator (that is, all terms on both sides of the equation):

$1x=-3$

$x=-3$

-3

Exercise #4

Solve the equation

$4\frac{1}{3}\cdot x=21\frac{2}{3}$

Step-by-Step Solution

We have an equation with a variable.

Usually, in these equations, we will be asked to find the value of the missing (X),

This is how we solve it:

To solve the exercise, first we have to change the mixed fractions to an improper fraction,

So that it will then be easier for us to solve them.

To convert a mixed fraction, we start by multiplying the whole number by the denominator

4*3=12

Now we add this to the existing numerator.

12+1=13

And we find that the first fraction is 13/3

Let's continue with the second fraction and do the same in it:
21*3=63

63+2=65

The second fraction is 65/3

We replace the new fractions we found in the equation:

13/3x = 65/3

At this point, we will notice that all the fractions in the exercise share the same denominator, 3.

Therefore, we can multiply the entire equation by 3.

13x=65

Now we want to isolate the unknown, the x.

Therefore, we divide both sides of the equation by the unknown coefficient -
13.

63:13=5

x=5

$x=5$

Exercise #5

$(7x+3)\times(10+4)=238$

Step-by-Step Solution

First, we solve the addition exercise in the right parenthesis:

$(7x+3)+14=238$

Now, we multiply each of the terms inside the parenthesis by 14:

$(14\times7x)+(14\times3)=238$

We solve each of the exercises inside the parenthesis:

$98x+42=238$

We move the sections and keep the appropriate sign:

$98x=238-42$

$98x=196$

We divide the two parts by 98:

$\frac{98}{98}x=\frac{196}{98}$

$x=2$

2

examples with solutions for first-degree equations

Exercise #1

$(a+3a)\times(5+2)=112$

Calculate a a

Step-by-Step Solution

First, we solve the two exercises in parentheses:

$4a\times7=112$

Divide each of the sections by 4:

$\frac{4a\times7}{4}=\frac{112}{4}$

In the fraction on the left side we simplify by 4 and in the fraction on the right side we divide by 4:

$a\times7=28$

Remember that:

$a\times7=a7$

Divide both sections by 7:

$\frac{a7}{7}=\frac{28}{7}$

$a=4$

4

Exercise #2

Solve for X:

$5x=25$

5

Exercise #3

Solve for X:

$\frac{1}{3}x=9$

27

Exercise #4

Solve for X:

$10+3x=19$

3

Exercise #5

Solve for X:

$24-8x=-2x$

4

examples with solutions for first-degree equations

Exercise #1

Find the value of the parameter X:

$x+5=8$

3

Exercise #2

Solve for X:

$5-x=4$

1

Exercise #3

Solve for X:

$6-4x=-18$

6

Exercise #4

Solve for X:

$33x-11x=66$

3

Exercise #5

Solve for X:

$-8x+3=-29$