**In this article we will get to know the equations and learn simple ways to solve them.**

## We will now look at equations with only one unknown

### For example

**Let's go back to the equation of the previous example:**

$X-1 = 5$

We want to isolate $X$. To do this we will add $1$ to both members of the equation.

**We will write it like this:**

$X-1 = 5$

**We will obtain:**

$x-1 + 1 = 5 + 1$

**That is:**

$X = 6$

And, this is the solution to our equation. We can always check if we got it right by putting our answer in the original equation. Let's put $X=6$ into the equation

$X-1 = 5$

and we get

$6-1 = 5$

$5=5$

this is a true statement, $5$ really equals $5$, i.e., our solution is correct.

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### Another example

$Z + 7 = 15$

First let's see that this time the variable is $Z$. The variable can be denoted by any letter we want.

As we have explained before, we are interested in finding the value of the $Z$ that will give us the solution to the equation. Therefore, we will now try to isolate $Z$. We will do this by subtracting $7$ from the two members of the equation.

It looks like this:

$Z + 7 = 15$

**We will obtain:**

$Z + 7 -7 = 15 โ 7$

$Z = 8$

This is the solution of the equation. Again, it is always convenient to check if we have found the correct value of the unknown by placing our answer in the original equation.

**Let's remember what the original equation was:**

$Z + 7 = 15$

let's put

$Z = 8$

**and we will get:**

$8 + 7 = 15$

$15=15$

This really is a true statement, i.e., the answer we received is correct.

## Solving equations by applying multiplication and division operations

So far we have solved equations by applying addition and subtraction operations to both sides of the equation. Now we will see other examples of equations that we **will solve with** **multiplication**** and division** **operations**:

Do you know what the answer is?

### Exercise 1

**Find the value of the unknown in the following equation and check that it is correct.**

$2X = 8$

**We want to isolate the** **$X$****.** **We divide both members of the equation by** **$2$****. We will write it like this:**

$2X/2= 8/2$

**and we will get :**

$X = 4$

**Also in this case it is convenient to place the solution in the original equation to see if we have done it right:**

$2\times 4 = 8$

$8 = 8$

We obtained a correct result, that is, our solution is right.

### Exercise 2

**Find the value of the unknown in the following equation and check that it is correct.**

$-3Y = 18$

To isolate the variable Y we divide both members of the equation by $-3$

$-3Y/-3 = 18/-3$

$Y = -6$

To verify our result, it is always convenient to place it in the original equation. Try it!

### Exercise 3

**Find the value of the unknown in the following equation and check that it is correct.**

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

Here we have a fraction in the equation. We want to get rid of it and isolate the $X$. We multiply both members of the equation by. $3$

$3\times\frac{1}{3}x=3\times5$

**We will obtain:**

$x=15$

**To verify this we will place the result obtained in the original equation:**

$\frac{1}{3}\times 15=5$

$5=5$

That is, the result obtained is correct.

### Exercise 4

**Find the value of the unknown in the following equation and check that it is correct.**

$2x + 3 = 5$

**This exercise requires subtracting and dividing operations. First, we subtract** **$3$**** from the two members of the equation:**

$2x + 3 = 5$

$2x + 3 - 3 = 5 โ 3$

$2x = 2$

**Now we will divide the two members of the equation by**. **$2$**** and we obtain:**

$2x/2 = 2/2$

$X = 1$

**Let's place the result obtained in the original equation to check if we have done it right:**

$2\times 1 + 3 = 5$

$5 = 5$

That is, the result obtained is correct.

Do you think you will be able to solve it?

### Exercise 5

**Find the value of the unknown in the following equation and check that it is correct.**

$โโโโโโโX-6=0$

**This exercise requires the operation of adding** **$6$**** in both members of the equation, so we have:**

$โโโโโโโX-6+6=0+6$

Simplifying we obtain that the solution of the equation is $X=6$ since if we put $6$ instead of the $X$ we will obtain the result $0$ on both sides of the equation, we will have two equivalent members.

### Exercise 6

**Find the value of the unknown in the following equation and check that it is correct.**

$2X-6=0$

**This exercise requires the operation of adding** **$6$**** in both members of the equation, so we have:**

$2X-6+6=0+6$

$2X=6$

**Now we divide by** **$2$**** both sides of the equation :**

$2X/2=6/2$

$X=3$

The solution of the equation is $X=3$ because if we put $3$ instead of the $X$ we will get the result $0$ on both sides of the equation, we will have two equivalent members.

### Exercise 7

**Find the value of the unknown in the following equation and check that it is correct.**

$3X-5=16$

**This exercise requires the operation of adding** **$5$**** in both members of the equation, so we have:**

$3X-5+5=16+5$

$3X=21$

**Now we divide by** **$3$**** both sides of the equation:**

$3X/3=21/3$

$X=7$

The solution of the equation is $X=7$ since if we put $7$ instead of the $X$ we will get the result $16$ on both sides of the equation, we will have two equivalent members.

## Questions on the subject

### How to clear an unknown?

Isolating the variable with mathematical operations.

Do you know what the answer is?

### How to isolate a variable or unknown?

Passing like terms to each side of the equality and performing mathematical operations.

### How to corroborate the solution of an equation?

Substituting the value found in the original equation and check that the equality is satisfied.

### What is an unknown?

It is the unknown value of the equation.

### How to solve a first order equation with one unknown?

Isolating the variable with mathematical operations.

Do you think you will be able to solve it?

### What is a first order equation with one unknown?

It is a mathematical equality involving a variable raised to the first power and fixed values that are numbers.

**If this article interested you, you may also be interested in the following articles:**

**In the** **Tutorela**** Math Blog you will find a wide variety of articles about mathematics**.

## examples with solutions for first-degree equations

### Exercise #1

Solve the equation

$20:4x=5$

### Video Solution

### 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

### Answer

### Exercise #2

Solve the equation

$5x-15=30$

### Video Solution

### 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

### Answer

### Exercise #3

Find the value of the parameter X

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

### Video Solution

### 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):

$1x=-3$

$x=-3$

### Answer

### Exercise #4

Solve the equation

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

### Video Solution

### 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.

Let's start with the four and the third:

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

### Answer

### Exercise #5

### Video Solution

### 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:

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

$x=2$

### Answer