Risolvere un'equazione di primo grado
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
📚 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
💡Variable
💡First member
💡Second member
💡Like terms
💡Coefficient
💡Multiplication
💡Equality sign
💡Solving
💡Terms
💡Parentheses
💡Simplifying
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
vogliamo risolvere questa equazione di
primo grado
ricordiamo che un equazione è
l'uguaglianza tra due espressioni di cui
almeno una letterale che risulta
verificata solo per particolare i valori
attribuiti alle incognite che in essa
compaiono nel nostro caso
l'equazione è un'equazione nella sola
incognita hicks tutta la parte che c'è a
sinistra dell'uguale è il primo membro
tutta la parte che c'è a destra del
segno di uguaglianza è il secondo membro
della nostra equazione vediamo un po
come si risolve un'equazione di primo
grado
allora la prima cosa da fare svolgere i
calcoli nei due membri
dopodiché separatamente prima secondo
membro si riducono i termini simili
quindi si portano tutti i termini con
l'aics per in generale con l'incognita a
primo membro e tutti i termini noti cioè
senza incognita a secondo membro si
riducono nuovamente i termini simili
eventualmente presenti e infine si
dividono entrambi membri per il
coefficiente dell'aics allora facciamo
passo passo la risoluzione della nostra
equazione osserviamo abbiamo due
parentesi tonde però dentro le parentesi
non ci sono termini simili
qui abbiamo un termine con l'aics un
termine senza e qui la stessa cosa
quindi non possiamo sommare nulla
all'interno delle parentesi andremo a
effettuare la moltiplicazione 4 per la
prima parentesi 2 per la seconda
parentesi guardiamo la prima come si fa
questa moltiplicazione bisogna
moltiplicare il termine che c'è
all'esterno separatamente per ciascuno
dei due termini contenuti nella
parentesi quindi 4x2 hicks
il risultato mi dà 8 x4 per meno due e
il risultato è meno 8
ripetiamo il segno di uguaglianza
facciamo quest'altra moltiplicazione 2 x
ics mida ii x2 per meno uno meno due e
riportiamo il più il che era fuori dalla
parentesi a questo punto abbiamo
effettuato il primo passaggio abbiamo
eseguito i calcoli nei due membri
dobbiamo ridurre ora i termini simili
notiamo che nel primo membro non ci sono
termini simili perché questo è un
termine che ha la ics e questo non ce
l'ha
nel secondo termine invece nel secondo
membro abbiamo 2x e
entrambi con parte letterale hicks
quindi si possono sommare allora il
primo membro lo dobbiamo riscrivere tale
quale non possiamo fare nulla
nel secondo 2x più hicks la somma è 3x
riportiamo il meno due
passiamo allora al punto 3 dobbiamo
portare i termini con l'aics a primo
membro i termini noti a secondo membro
per fare questo ci dobbiamo ricordare
che quando un termine cambiali posto
rispetto al quale cioè passa dal primo
al secondo membro viceversa deve
cambiare di segno allora 8x lo
riscriviamo uguale perché sta già primo
membro 3x lo vogliamo portare da destra
a sinistra quindi diventerà meno 3x
perché deve cambiare segno uguale questo
numero meno 8 è un termine noto quindi
lo devo portare al secondo me
cambiandogli disegno diventa più 8
meno due si trova già dalla parte giusta
e quindi lo riscrivo esattamente uguale
dobbiamo ora ridurre di nuovo i termini
simili siamo al punto 4 nel primo membro
entrambi hanno parte letterale hicks
quindi li sommiamo ottenendo 5x
ma secondo membro sono termini noti
quindi si possono sommare 8 meno 26
siamo arrivati così all'ultimo punto
dobbiamo dividere entrambi membri per il
coefficiente dell'aics che nel nostro
caso e 5a primo membro 55 si semplifica
noi otteniamo quindi che hicks uguale a
6
quindi è la soluzione della nostra
equazione di primo grado
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