Risolvere un'equazione di primo grado

HUB Scuola
21 Jul 201903:53

Summary

TLDRThe video script offers a detailed walkthrough of solving a first-degree equation, specifically focusing on the variable Hicks. It emphasizes the importance of performing calculations on both sides of the equation, combining like terms, and isolating the variable. The process involves multiplying terms outside parentheses by those inside, moving terms with the variable to one side, and those without to the other, changing the sign as they move. The script concludes with dividing both sides by the coefficient of the variable to find its value. In this case, the solution to the equation is Hicks equals 6, providing a clear example of solving a linear equation step by step.

Takeaways

  • 📚 The script discusses solving a first-degree equation, emphasizing the importance of understanding the structure of an equation with two expressions separated by an equals sign.
  • 🔍 It defines the first member (left side of the equals sign) and the second member (right side) of the equation, highlighting that at least one of them contains a variable.
  • ✅ The process of solving the equation involves performing calculations on both sides, simplifying terms, and isolating the variable.
  • 🧮 It is mentioned that similar terms on each side of the equation should be combined to simplify the equation further.
  • 🤔 The script provides a step-by-step guide on how to solve the equation, starting with performing calculations on both sides.
  • 📝 It explains the need to move terms with the variable to one side of the equation and constants to the other, changing their signs in the process.
  • 📊 The multiplication of terms outside parentheses with those inside is demonstrated, emphasizing the distributive property.
  • 🧐 The transcript illustrates the process of combining like terms on both sides of the equation after performing the multiplication.
  • 🔄 It is clarified that when moving terms from one side of the equation to the other, their signs must be reversed.
  • ➗ Finally, the script describes dividing both sides of the equation by the coefficient of the variable to solve for its value.
  • 🎓 The solution to the given first-degree equation is found to be hicks equals 6, after following the outlined steps.
  • 📈 The script serves as a tutorial for individuals learning to solve linear equations, providing a clear methodological approach.

Q & A

  • What is the main topic of the transcript?

    -The main topic of the transcript is solving a first-degree equation, specifically focusing on the steps to resolve it.

  • What is the first step in solving a first-degree equation according to the transcript?

    -The first step is to perform calculations in both the left and right sides of the equation.

  • What does the transcript refer to as the 'first member' and 'second member' of the equation?

    -The 'first member' refers to the part of the equation to the left of the equals sign, and the 'second member' refers to the part to the right of the equals sign.

  • How does the transcript describe the process of simplifying terms in the equation?

    -The transcript describes simplifying terms by reducing similar terms separately in the first and second members and then moving all terms with the unknown variable to the first member and all known terms to the second member.

  • What is the purpose of moving terms from one side of the equation to the other?

    -The purpose is to isolate the variable on one side of the equation, which is a step towards finding its value.

  • What is the significance of changing the sign when moving terms from one side of the equation to the other?

    -Changing the sign is necessary to maintain the balance of the equation, as the term's contribution to the equation's sum changes when it is moved.

  • What does the transcript indicate as the final step in solving the equation?

    -The final step is to divide both members of the equation by the coefficient of the unknown variable to find its value.

  • What is the solution to the first-degree equation presented in the transcript?

    -The solution to the equation is hicks equals 6.

  • How does the transcript handle terms within parentheses during the equation-solving process?

    -The transcript advises to multiply the term outside the parentheses by each term inside the parentheses, taking care to distribute the multiplication correctly.

  • What is the role of the unknown variable in the context of the transcript?

    -The unknown variable, represented as 'hicks' in the transcript, is the element of the equation that the process aims to solve for, by isolating it and determining its value.

  • Why is it important to combine like terms during the equation-solving process?

    -Combining like terms simplifies the equation and makes it easier to solve by reducing the number of variables and constants that need to be dealt with separately.

  • How does the transcript ensure that the equation remains balanced during the solving process?

    -The transcript ensures balance by applying the same operations to both sides of the equation and by changing the sign of terms when they are moved from one side to the other.

Outlines

00:00

📚 Introduction to Solving First Degree Equations

This paragraph introduces the concept of a first degree equation, explaining that it involves an equality between two expressions with at least one variable. It defines the first and second members of the equation and sets the stage for solving the given equation step by step.

🧮 Performing Calculations in the Equation

The paragraph outlines the first step in solving the equation, which is to perform calculations in both the first and second members. It emphasizes the need to simplify terms and combine like terms, ultimately resulting in a simplified form of the equation.

🔍 Distributing Terms and Simplifying

This section focuses on the process of distributing terms within parentheses and simplifying the equation further. It demonstrates the multiplication of terms outside the parentheses with those inside, followed by combining like terms and ensuring the equation is properly simplified.

🔄 Rearranging Terms and Changing Signs

The paragraph explains the next step of moving terms with the variable (hicks) to the first member and constants to the second member. It highlights the importance of changing the sign of terms when they are moved from one side of the equation to the other.

🧐 Combining Like Terms and Simplifying Further

This section discusses the process of combining like terms in both the first and second members of the equation. It shows how to sum terms with the variable and constants separately, resulting in a simpler form of the equation.

🏁 Final Step: Dividing by the Variable Coefficient

The final paragraph describes the last step in solving the equation, which is to divide both members by the coefficient of the variable (5 in this case). This simplifies the equation to find the value of hicks, which is determined to be 6, providing the solution to the first degree equation.

Mindmap

Keywords

💡First-degree equation

A first-degree equation, also known as a linear equation, involves a single variable raised to the power of one. It is the simplest form of a polynomial equation and is central to the video's theme of solving equations. In the script, the equation is described as having two expressions separated by an equality sign, with at least one variable that finds specific values to satisfy the equation.

💡Variable

A variable is a symbol, often a letter, that represents an unknown quantity in an equation. In the context of the video, 'hicks' is the variable in the first-degree equation that the speaker is solving. The process of solving the equation involves finding the value of 'hicks' that makes the equation true.

💡First member

The first member of an equation refers to the expression that appears to the left of the equality sign. In the video script, the first member includes terms with the variable 'hicks' before any operations are performed. It is a crucial part of setting up the equation and eventually finding the solution.

💡Second member

The second member is the expression on the right side of the equality sign in an equation. It typically contains constant terms or terms without the variable. In the script, the second member is manipulated to isolate the variable on one side of the equation, which aids in solving for its value.

💡Like terms

Like terms are terms in a mathematical expression that contain the same variables raised to the same power. The video emphasizes combining like terms as part of the equation-solving process. For instance, '2x' and 'hicks' are like terms in the script that can be combined to simplify the equation.

💡Coefficient

A coefficient is a numerical factor that multiplies a variable in an equation. In the context of the video, the coefficient of 'hicks' is an important part of the final step, where both sides of the equation are divided by this coefficient to isolate the variable and find its value.

💡Multiplication

Multiplication is a mathematical operation that combines two numbers to make a product. In the script, multiplication is used to distribute terms outside parentheses across the terms inside, such as '4 times hicks' resulting in '8 hicks', which is a step in simplifying the equation.

💡Equality sign

The equality sign (=) is a symbol used to indicate that two expressions are equivalent. It is a fundamental part of any equation and is used repeatedly in the script to maintain the balance of the equation as terms are moved from one side to the other.

💡Solving

Solving refers to the process of finding the value or values of the variable that satisfy the equation. The entire narrative of the video is centered around solving the first-degree equation for the variable 'hicks', which involves several steps including combining like terms and isolating the variable.

💡Terms

In the context of an equation, terms are the individual components that make up the expressions on each side of the equality sign. The script discusses terms with variables (like 'hicks') and terms without variables (constants), which are manipulated to find the solution to the equation.

💡Parentheses

Parentheses are symbols used in mathematical expressions to group or encapsulate terms that should be evaluated together before operations outside the parentheses are performed. In the script, parentheses are used to indicate that the terms inside should be multiplied by the numbers outside before proceeding with the rest of the equation.

💡Simplifying

Simplifying an equation involves reducing it to a more straightforward form by combining like terms, performing arithmetic operations, and isolating the variable. The script demonstrates this process step by step, showing how the initial complex equation is broken down to find the value of 'hicks'.

Highlights

The transcript discusses the process of solving a first-degree equation.

An equation is defined as an equality between two expressions with at least one variable.

The left side of the equation is referred to as the first member, while the right side is the second member.

The first step in solving an equation is to perform calculations separately in the two members.

After calculations, terms with similar variables are combined.

All terms with the variable are placed in the first member, and all constant terms in the second member.

Terms are further simplified by combining like terms, if present.

The final step is to divide both members by the coefficient of the variable to isolate it.

The example given involves an equation with the variable 'hicks'.

The process starts with expanding parentheses and multiplying terms outside by those inside.

After expanding, there are no like terms within the parentheses to combine.

In the second member, like terms (2x and x) are combined to form 3x.

Constants are moved to the right side of the equation by changing their sign.

The equation is then simplified by combining like terms in both members.

The final equation is obtained by dividing by the coefficient of 'hicks', which is 5.

The solution to the equation is found to be 'hicks' equals 6.

The method described can be applied to any first-degree equation with a single variable.

This process is a fundamental aspect of algebra and is essential for solving more complex equations.

Transcripts

play00:06

vogliamo risolvere questa equazione di

play00:09

primo grado

play00:10

ricordiamo che un equazione è

play00:12

l'uguaglianza tra due espressioni di cui

play00:15

almeno una letterale che risulta

play00:17

verificata solo per particolare i valori

play00:19

attribuiti alle incognite che in essa

play00:21

compaiono nel nostro caso

play00:23

l'equazione è un'equazione nella sola

play00:25

incognita hicks tutta la parte che c'è a

play00:29

sinistra dell'uguale è il primo membro

play00:31

tutta la parte che c'è a destra del

play00:33

segno di uguaglianza è il secondo membro

play00:35

della nostra equazione vediamo un po

play00:37

come si risolve un'equazione di primo

play00:40

grado

play00:41

allora la prima cosa da fare svolgere i

play00:43

calcoli nei due membri

play00:44

dopodiché separatamente prima secondo

play00:47

membro si riducono i termini simili

play00:48

quindi si portano tutti i termini con

play00:51

l'aics per in generale con l'incognita a

play00:53

primo membro e tutti i termini noti cioè

play00:55

senza incognita a secondo membro si

play00:58

riducono nuovamente i termini simili

play01:00

eventualmente presenti e infine si

play01:02

dividono entrambi membri per il

play01:04

coefficiente dell'aics allora facciamo

play01:06

passo passo la risoluzione della nostra

play01:09

equazione osserviamo abbiamo due

play01:11

parentesi tonde però dentro le parentesi

play01:13

non ci sono termini simili

play01:15

qui abbiamo un termine con l'aics un

play01:17

termine senza e qui la stessa cosa

play01:19

quindi non possiamo sommare nulla

play01:20

all'interno delle parentesi andremo a

play01:22

effettuare la moltiplicazione 4 per la

play01:25

prima parentesi 2 per la seconda

play01:27

parentesi guardiamo la prima come si fa

play01:30

questa moltiplicazione bisogna

play01:32

moltiplicare il termine che c'è

play01:33

all'esterno separatamente per ciascuno

play01:35

dei due termini contenuti nella

play01:36

parentesi quindi 4x2 hicks

play01:39

il risultato mi dà 8 x4 per meno due e

play01:44

il risultato è meno 8

play01:47

ripetiamo il segno di uguaglianza

play01:49

facciamo quest'altra moltiplicazione 2 x

play01:52

ics mida ii x2 per meno uno meno due e

play01:57

riportiamo il più il che era fuori dalla

play01:59

parentesi a questo punto abbiamo

play02:01

effettuato il primo passaggio abbiamo

play02:03

eseguito i calcoli nei due membri

play02:05

dobbiamo ridurre ora i termini simili

play02:08

notiamo che nel primo membro non ci sono

play02:10

termini simili perché questo è un

play02:11

termine che ha la ics e questo non ce

play02:13

l'ha

play02:14

nel secondo termine invece nel secondo

play02:16

membro abbiamo 2x e

play02:18

entrambi con parte letterale hicks

play02:21

quindi si possono sommare allora il

play02:23

primo membro lo dobbiamo riscrivere tale

play02:25

quale non possiamo fare nulla

play02:27

nel secondo 2x più hicks la somma è 3x

play02:31

riportiamo il meno due

play02:35

passiamo allora al punto 3 dobbiamo

play02:37

portare i termini con l'aics a primo

play02:39

membro i termini noti a secondo membro

play02:41

per fare questo ci dobbiamo ricordare

play02:43

che quando un termine cambiali posto

play02:45

rispetto al quale cioè passa dal primo

play02:47

al secondo membro viceversa deve

play02:49

cambiare di segno allora 8x lo

play02:52

riscriviamo uguale perché sta già primo

play02:54

membro 3x lo vogliamo portare da destra

play02:58

a sinistra quindi diventerà meno 3x

play03:00

perché deve cambiare segno uguale questo

play03:04

numero meno 8 è un termine noto quindi

play03:05

lo devo portare al secondo me

play03:07

cambiandogli disegno diventa più 8

play03:10

meno due si trova già dalla parte giusta

play03:12

e quindi lo riscrivo esattamente uguale

play03:15

dobbiamo ora ridurre di nuovo i termini

play03:17

simili siamo al punto 4 nel primo membro

play03:20

entrambi hanno parte letterale hicks

play03:22

quindi li sommiamo ottenendo 5x

play03:26

ma secondo membro sono termini noti

play03:27

quindi si possono sommare 8 meno 26

play03:31

siamo arrivati così all'ultimo punto

play03:33

dobbiamo dividere entrambi membri per il

play03:36

coefficiente dell'aics che nel nostro

play03:38

caso e 5a primo membro 55 si semplifica

play03:42

noi otteniamo quindi che hicks uguale a

play03:44

6

play03:45

quindi è la soluzione della nostra

play03:48

equazione di primo grado

Rate This

5.0 / 5 (0 votes)

Related Tags
Algebraic EquationFirst-DegreeSolving TechniquesMathematical ProcessEducational ContentStep-by-Step GuideEquation ManipulationSimplificationMath TutorialLearning ResourceMathematical Concepts