Solve a system with three variables
TLDRThe transcript outlines a step-by-step process for solving a system of three-variable linear equations. It emphasizes the importance of eliminating one variable to simplify the system and then solving the resulting two-variable system. The method involves strategic multiplication of equations and addition or subtraction to isolate variables. The example provided walks through the process of eliminating 'y', solving for 'x' and 'z', and finally substituting these values back into one of the original equations to find 'y'. The end result is the solution to the three-variable system.
Takeaways
- 📝 The script outlines a step-by-step process for solving a system of linear equations with three variables.
- 🔍 The first step involves eliminating one variable by using two pairs of equations, resulting in a system of two equations with two variables.
- 🎯 The variable to eliminate should be chosen based on ease of elimination, often picking one with a coefficient of 1.
- 🖋️ The process of elimination can be done by multiplying equations to make coefficients compatible for addition or subtraction.
- 📊 After eliminating the variable, two new equations (labeled as 'a' and 'b' in the script) are obtained, each with two variables.
- 🛠️ The second step is to solve the new system of two equations to find the values of the remaining two variables.
- 🔄 The third step involves substituting the found values back into one of the original equations to find the value of the third variable.
- 📐 The method used in the script demonstrates the use of both addition and subtraction for elimination, as well as multiplication to adjust coefficients.
- 🤔 The script emphasizes the importance of careful selection of which variable to eliminate and how to manipulate the equations for efficient solving.
- 👉 The process is demonstrated using a specific set of equations, but the method is applicable to any system of three-variable linear equations.
- 📝 The final solution to the example provided in the script is x = -1, y = -5, and z = 7.
Q & A
What is the main topic of the transcript?
-The main topic of the transcript is solving a system of three-variable linear equations using the process of elimination.
How many equations and variables are present in the system described in the transcript?
-There are three equations and three variables in the system: x, y, and z.
What is the first step in solving the system of equations as described in the transcript?
-The first step is to eliminate one variable by using two pairs of equations, creating a system of two equations with two variables.
Which variable does the speaker choose to eliminate first, and why?
-The speaker chooses to eliminate the variable y first because it has a coefficient of 1, making the process of elimination simpler.
How does the speaker eliminate the variable y from the first two equations?
-The speaker eliminates y by multiplying the second equation by 2 and adding it to the first equation, resulting in a new equation with no y term.
What are the two new equations obtained after the first elimination process?
-The two new equations obtained are: -5x - 6z = -37 (Equation A) and -4x + y - 5z = -36 (the unchanged second original equation).
How does the speaker proceed to eliminate y from the second and third equations?
-The speaker multiplies the second equation by 3 and adds it to the third equation to eliminate y.
What are the resulting equations after the second elimination process?
-The resulting equations are: -7x - 12z = -77 (Equation B) and 5x - 3y + 3z = 31 (the unchanged third original equation).
How does the speaker solve the system of two-variable equations (A and B)?
-The speaker multiplies Equation A by -2 to make the coefficients of z match, then adds it to Equation B, resulting in a single equation with only x and z. From this, the speaker finds that x = -1 and z = 7.
What is the final step in solving the original three-variable system?
-The final step is to substitute the values of x and z back into one of the original equations to solve for y. The speaker substitutes these values into the first original equation, resulting in y = -5.
What are the solutions for the variables x, y, and z in the system?
-The solutions for the variables are x = -1, y = -5, and z = 7.
Outlines
📚 Introduction to Solving a Three-Variable System
The speaker begins by explaining that the video will cover the process of solving a three-variable system of equations. They emphasize that although it might seem complex at first, the audience will grasp the concept more easily as the lesson progresses. The system given consists of three equations with three variables: 3x - 2y + 4z = 35, -4x + y - 5z = -36, and 5x - 3y + 3z = 31. The initial step is to eliminate one variable by using two pairs of equations, which simplifies the system to a more manageable two-variable system. The speaker suggests that the easiest variable to eliminate is the one with a coefficient of 1, which in this case is 'y'. They proceed to demonstrate the elimination process by multiplying the second equation by 2 to align coefficients and then adding it to the first equation to eliminate 'y', resulting in a new equation labeled as 'equation a'.
🔢 Elimination of 'y' and Creation of Equations A and B
Continuing from the previous explanation, the speaker focuses on the elimination of the 'y' variable from the entire system. They reiterate the importance of consistently targeting 'y' for elimination across all equations. By multiplying the top equation by 3 and the bottom equation by an appropriate factor, they create a new equation 'b'. The speaker then adds equations A and B, using the elimination technique, to find a relationship between 'x' and 'z'. However, they make a mistake in the process and correct it by multiplying equation A by -2 instead of 2, which results in a new equation with the correct signs for further operations. The speaker emphasizes the importance of minimizing errors during the elimination process, suggesting multiplication by a negative number as a safer approach.
🎓 Solving for 'x' and 'z', and Back-substitution
In this segment, the speaker successfully solves for 'x' and 'z' using the two newly created equations, A and B. They add the equations, which have been adjusted to have the same terms, to eliminate 'z' and find the value of 'x'. After solving, they obtain x = -1. The speaker then substitutes the value of 'x' back into one of the original equations to solve for 'z', resulting in z = 7. With the values of 'x' and 'z' determined, the speaker moves on to step three, which involves substituting these values into one of the original equations to solve for the remaining variable 'y'. After a brief calculation, the speaker finds y = -5. The speaker concludes by presenting the solution to the three-variable system as x = -1, y = -5, and z = 7, emphasizing the manageability of the process despite its initial complexity.
Mindmap
Keywords
💡transcript
💡system of equations
💡variables
💡elimination
💡substitution
💡coefficients
💡linear equations
💡solving for
💡algebra
💡mathematical process
Highlights
The process begins by emphasizing the importance of patience and persistence when solving complex mathematical problems, particularly those involving multiple variables.
The transcript introduces a three-variable system with three equations, highlighting the significance of having an equal number of variables and equations for effective problem-solving.
The concept of elimination is discussed as a key strategy for simplifying the system by using pairs of equations to remove one variable at a time.
The choice of which variable to eliminate is crucial and should be based on the coefficients of the variables, preferably starting with one that has a coefficient of 1 for ease of computation.
The transcript provides a step-by-step guide on how to eliminate the variable 'y' by multiplying the second equation by 2 and then adding it to the first equation.
After eliminating 'y', the resulting equation is labeled as 'equation a', demonstrating the importance of organization and clear labeling in solving complex problems.
The process is repeated to eliminate 'y' between the second and third equations, resulting in a new equation 'b', further simplifying the system.
The transcript emphasizes the use of tips and strategies, such as multiplying by negative numbers to avoid subtraction, which can lead to errors.
Once equations 'a' and 'b' are established, the next step is to solve this system of two equations to find the values of the remaining two variables, 'x' and 'z'.
The solution process involves multiplying equations to make the coefficients of one variable the same, which simplifies the elimination process.
The value of 'x' is determined to be negative one, and 'z' is found to be seven, demonstrating the successful application of the elimination method.
The final step involves substituting the found values of 'x' and 'z' back into one of the original equations to solve for the remaining variable 'y'.
Through the substitution process, 'y' is found to be negative five, completing the solution of the three-variable system.
The transcript concludes by summarizing the entire process, emphasizing the methodical approach to solving complex mathematical problems and the importance of each step.
The importance of patience, clear labeling, and strategic variable elimination is reiterated, providing a comprehensive guide for solving similar problems in the future.
The transcript serves as a practical guide for solving systems of equations with three variables, showcasing the application of mathematical strategies and techniques.
Transcripts
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