The solution of an equation is, in fact, a numerical value that, if we place it in place of the unknown (or the variable), we will achieve equality between the two members of the equation, that is, we will obtain a "true statement". In first degree equations with one unknown, there can only be one solution.

Example:

$X - 1 = 5$

This is an equation with one unknown or variable indicated by the letter $X$. The equation is composed of two members separated by the use of the equal sign $=$. The left member is everything to the left of the sign $=$, and the right member is everything to the right of the sign.

Our goal is to isolate the variable (or clear the variable) $X$ so that only it remains in one of the members of the equation. In this way we will discover its value. In this article we will learn how to use the four mathematical operations(addition, subtraction, multiplication and division) to isolate the variable $X$.

examples with solutions for linear equations

Exercise #1

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 #2

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 #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

Given: the length of a rectangle is 3 greater than its width.

The area of the rectangle is equal to 27 cm².

Calculate the length of the rectangle

Step-by-Step Solution

The area of the rectangle is equal to length multiplied by width.

Let's set up the data in the formula:

$27=3x\times x$

$27=3x^2$

$\frac{27}{3}=\frac{3x^2}{3}$

$9=x^2$

$x=\sqrt{9}=3$

$x=3$

Exercise #6

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

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

Exercise #7

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

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 #8

$4x:30=2$

Video Solution

$x=15$

Exercise #9

Solve the equation

$7x+5.5=19.5$

Video Solution

$x=2$

Exercise #10

Solve the equation

$8x\cdot10=80$

Video Solution

$x=1$

Exercise #11

$14x+3=17$

Video Solution

$x=1$

Exercise #12

$5x=0$

Video Solution

$x=0$

Exercise #13

$5x=1$

What is the value of x?

Video Solution

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

Exercise #14

Find the value of the parameter X:

$x+5=8$

3

Exercise #15

Solve for X:

$3+x=4$