Equations. First degree equation

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First degree equation

What is an equation?

An equation in mathematics is a statement that links 2 expressions with the equal symbol.
One of the basic type of equations are the first degree equations,that can be called as ‘’Linear equations”.

Linear Equations with one variable

A linear equation is an equation that represents a line on the quadratic system.
The general form of this equation is ax + b = 0 in which a and b are integers and x is the variable. This type of equation has only one solution and represents a parallel line with the y-axis.

How to solve linear equations with one variable?

Firstly let’s write the basic Axioms that we apply when solving an equation:

1. Addition axiomWhen two equal quantities are added on the both sides of an equation, the equation will still remain equal.

2. Subtraction axiomWhen two equal quantities are subtracted on the both sides of an equation the equation will still remain equal.

3. Multiplication axiom: When we multiply the both sides of an equation with the same value, the equation will still remain equal.

4. Division axiom: When we divide both sides of an equation with the same value(≠0), the equation will still remain equal.

5. Distributive axioma(b+c) = ab + ac.

Steps for solving a linear equation with one variable: ax+b=0

What we have to do is figure out how to isolate the variable x and for that the axioms above will help us. As a result, we will be using the axioms depending on the equation that we have.

1) Firstly we have to see which is the variable that we need to isolate.

2) Then distinguish the variables and the constants.

3) Group the variables on the left side and the constants on the right.

4) Using axioms that we listed above we perform algebraic operations so we can get the value of the variable.

Worked example 1: Solve 6x + 8 = 12

SolutionOur aim is to isolate the variable x.

Step 1Substract 8 to both sides.

6x + 8 – 8 = 12 – 8

6x = 4

Step 2Divide both sides with 6

$\displaystyle \frac{6x}{x}=\frac{4}{6}$

$\displaystyle x=\frac{2}{3}$

Step 3So we isolated the variable $\displaystyle x=\frac{2}{3}$

Worked example 2Solve 3(x+8) – 2 = 3(9-x)

Solution: Firstly we need to apply the distrubutive axiom or simply said remove the parathenses.

3(x+8) – 2 = 3(9-x)

3x + 24 – 2 = 27 – 3x     (calculate the like terms on both sides, if we have)

3x + 22 = 27 – 3x

Secondly, we should Combine the like terms.

We need to place the ‘x’ on the same side (left side preferred) and constants on the other side, right side.

After that, we add 3x to both sides and subtract 22 from both sides.

3x + 22 + 3x = 27 – 3x + 3x

6x + 22 = 27

6x + 22 – 22 = 27 – 22

6x = 5

Thirdly, all it’s left to do is divide both sides by 6 so we can isolate the ‘x’.

$\displaystyle \frac{6x}{6}=\frac{5}{6}$

$\displaystyle x=\frac{5}{6}$

How to solve first degree equation with fractions.

First degree equations with fractions are equations that are a little bit more difficult to solve.

So you need to follow some steps carefully to avoid mistakes.

1. Firstly we need to remove the denominator.
2. Then remove the parenthesis.
3. Relocate the terms, where the variables to the left side and numbers to the right side of the equation.
4. Simplify by doing mathematical operations.
5. Find the value of x.

But how can we remove a denominator in a first degree equation?

  • Firstly we have to obtain the common denominator of all the denominators of the equation in order to add and subtract fractions.

  • After finding the common denominator we multiply the numerator by its corresponding number to obtain its equivalents fractions.

  • This number is obtained by dividing the common denominator by the denominator of the  original fraction.

  • After this we can eliminate the denominator in both sides and then we follow the steps as above.

Example 1: Solve the equation:  $\displaystyle \frac{{6x+2}}{3}-1=3x$

Step 1$\displaystyle \frac{?}{3}-\frac{3}{3}=\frac{?}{3}$

$\displaystyle \frac{{(6x+2)}}{3}-\frac{3}{3}=\frac{{3\cdot 3x}}{3}$

$\displaystyle (6x+2)-3=9x$

Step 2$\displaystyle 6x+2-3=9x$

Step 3$\displaystyle 6x-9x=+3-2$

Step 4$\displaystyle -3x=1$

Step 5$\displaystyle x=-\frac{1}{3}$

Example 2: Solve the equation:$\displaystyle 2x+1-\frac{{x+1}}{4}=\frac{x}{3}$

1. $\displaystyle \frac{?}{{12}}+\frac{?}{{12}}-\frac{?}{{12}}=\frac{?}{{12}}$

$\displaystyle \frac{{12(2x)}}{{12}}+\frac{{12}}{{12}}-\frac{{3(x+1)}}{{12}}=\frac{{4x}}{{12}}$

$\displaystyle 12(2x)+12-3(x+1)=4x$

2. $\displaystyle 24x+12+-3x-3=4x$

3. $\displaystyle 24x-3x-4x=3-12$

4. $\displaystyle 17x=-9$

5. $\displaystyle x=-\frac{9}{{17}}$

Solving problems by using a first degree equations with one variable

Step 1: Firstly we underline the key data of the problem.

Step 2: Then we determine the variable that is always what the problem is asking.

Step 3: Make the connection between the variable and the other data of the problem.

Step 4: Then we write our problem in form of an equation.

Step 5: We solve it based on the steps as above.

Step 6: The proof

Problem 1: A book with the cover costs 24 dollars. Only the cover costs 20% of the value of the book without a cover. How much does it costs the book without a cover?

SolutionThe book with the cover costs 24 dollars.

The cover costs 20%  the value of the book without a cover

Mark the Variable
The problem is asking how much the book without a cover costs so we mark that as our variable.
X = how much does the book without a cover costs.

Now, we will make the connection between the variable and the key data.
Book without the cover: X
The cover: 20% of X

We write our problem in form of an equation.
$\displaystyle x+\frac{20}{100}x=24$

$\displaystyle x+\frac{1}{5}x=24$

 We use the multiplication axiom to get rid of the fraction.

$\displaystyle 5x+5\frac{1}{5}x=24$

$\displaystyle 5x+5\frac{1}{5}x=5\cdot 24$

$\displaystyle 5x+x=120$

$\displaystyle 6x=120$

$\displaystyle x=\frac{120}{6}$
x = 20

The answer: The book without a cover costs 20 dollars.

The proof:
The cover costs 20% of 20 dollars = 4 dollars.
20 dollars + 4 dollars = 24 dollars that costs the book.

See also: Quadratic equation

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