Substitution Method For Solving Systems of Linear Equations, 2 and 3 Variables, Algebra 2
TLDRThis instructional video demonstrates the substitution method for solving systems of linear equations. It begins with a simple example involving two equations with two variables, guiding viewers through the process of substitution and simplification to find the values of x and y. The video then tackles more complex scenarios, including equations with fractions and multiple variables, showing how to eliminate fractions and systematically solve for each variable. Each example is followed by a step-by-step explanation, making the content accessible for viewers to understand and apply the substitution method to various problems.
Takeaways
- 📚 The video focuses on solving systems of linear equations using the substitution method.
- 🔍 The first example involves two equations with 'y' expressed in terms of 'x', which are set equal to each other to find a single variable equation.
- 📝 In the first example, by substituting and simplifying, it's found that x = 3 and subsequently y = 14, resulting in the solution (3,14).
- 📈 The second example deals with equations involving fractions, where the least common multiple (LCM) is used to eliminate the fractions before solving.
- 🧩 For the fraction-based example, after clearing the fractions and simplifying, it's determined that x = 6 and y = 0, leading to the solution (6,0).
- 🔑 The third example provides a system with two equations and demonstrates substituting 'y' to find x = 2 and y = 3, with the solution (2,3).
- 🔄 In the fourth example, 'y' is again substituted into the first equation, resulting in x = 4 and y = 3, giving the solution (4,3).
- 📐 The final example increases complexity with three equations and variables, where 'z' and 'y' are substituted in terms of 'x' to find x = 1, y = 2, and z = 3, resulting in (1,2,3).
- 📘 The video script provides step-by-step instructions on how to substitute and solve for variables in systems of linear equations.
- 👍 The method demonstrated is effective for solving systems of equations, whether they involve two or three variables, and with or without fractions.
Q & A
What is the substitution method for solving a system of linear equations?
-The substitution method involves replacing one variable in a set of equations with an expression that represents its value in terms of the other variables, allowing you to solve for one variable at a time.
How do you find the value of x and y from the equations y = 2x + 8 and y = 5x - 1?
-You set the two expressions for y equal to each other (5x - 1 = 2x + 8) and solve for x. Once x is found, you substitute it back into either original equation to find the value of y.
What is the solution to the system of equations y = 2x + 8 and y = 5x - 1?
-The solution is the ordered pair (x, y) = (3, 14), found by substituting and solving the equations as described.
How do you handle fractions when using the substitution method?
-You should eliminate fractions by finding the least common multiple (LCM) of the denominators and multiplying every term in the equation by this LCM to clear the fractions.
What is the least common multiple (LCM) between 2, 3, and 4, and why is it used?
-The LCM between 2, 3, and 4 is 12. It is used to eliminate fractions by ensuring that every term in the equation can be multiplied by this number without leaving any fractions.
How do you solve the equation 1/4x - 3/2 = 2/3x - 4 after clearing the fractions?
-After multiplying through by 12 to clear the fractions, you get 3x - 18 = 8x - 48. You then solve for x by combining like terms and isolating x on one side of the equation.
What is the solution to the system of equations y = 2/3x - 4 and y = 1/4x - 3/2?
-The solution is the ordered pair (x, y) = (6, 0), found by substituting and solving the equations after clearing the fractions.
How do you approach a system of equations with three variables?
-You can use the substitution method by expressing two of the variables in terms of the third and then solving the resulting equations step by step.
What is the solution to the system of equations x + y + z = 6, z = x + 2, and y = 2x?
-The solution is the ordered triple (x, y, z) = (1, 2, 3), found by substituting the expressions for y and z into the first equation and solving for x, then finding y and z.
How do you solve the system of equations 5x - 4y = 8 and y = 1/2x + 1?
-You substitute y in the first equation with 1/2x + 1, then solve for x. After finding x, you substitute it back into the equation for y to find its value.
What is the solution to the system of equations 5x - 4y = 8 and y = 1/2x + 1?
-The solution is the ordered pair (x, y) = (4, 3), found by substituting and solving the equations as described.
Outlines
📚 Introduction to Solving Systems of Linear Equations by Substitution
This paragraph introduces the concept of solving systems of linear equations using the substitution method. The video begins with two equations involving 'x' and 'y', and demonstrates the process of substitution by replacing 'y' with its equivalent expression from the second equation. The goal is to simplify the system to a single equation with one variable, which can then be solved. The example provided involves basic algebraic manipulation, such as adding and subtracting terms, and dividing by a coefficient to isolate the variable. The solution for 'x' is found, and then substituted back into one of the original equations to find the corresponding value of 'y'. The final answer is presented as an ordered pair (x, y).
🔍 Solving Systems with Fractions and Multiple Examples
The second paragraph continues the theme of substitution but introduces fractions into the equations. The video script guides through the process of eliminating fractions by finding the least common multiple (LCM) of the denominators, which in this case is 12. Each term of the equation is then multiplied by 12 to clear the fractions. The resulting equation is simplified by combining like terms and isolating the variable 'x'. The solution for 'x' is found and substituted back into one of the original equations to solve for 'y'. The process is repeated with additional examples, each time emphasizing the substitution method and the algebraic steps required to find the solution for the variables involved. The solutions are presented as ordered pairs, demonstrating the successful application of the substitution method.
📘 Advanced Substitution with Three Variables
The final paragraph of the script tackles a more complex system involving three variables (x, y, z). The approach remains the substitution method, but with an additional variable, the process requires careful tracking of each substitution. The script shows how to replace 'z' and 'y' in the first equation with their respective expressions in terms of 'x'. This substitution leads to an equation with a single variable, which is then solved for 'x'. Once 'x' is determined, it is used to find the values of 'y' and 'z' by substituting back into their respective equations. The solution to this system is presented as an ordered triplet (x, y, z), illustrating the successful application of the substitution method to a more complex scenario.
Mindmap
Keywords
💡Substitution method
💡System of linear equations
💡Variable
💡Least common multiple (LCM)
💡Ordered pair
💡Distribute
💡Combine like terms
💡Equation
💡Fraction
💡Solve
Highlights
Introduction to solving a system of linear equations using the substitution method.
Problem setup with two equations involving y: y = 2x + 8 and y = 5x - 1.
Substitution of y with 5x - 1 to create a single equation with one variable.
Solving the equation 5x - 1 = 2x + 8 by simplifying and isolating x.
Finding x = 3 by solving the simplified equation 3x = 9.
Substituting x = 3 back into the original equations to find y.
Calculation of y = 14 using the value of x.
Presenting the solution as an ordered pair (x, y) = (3, 14).
Introduction to a second example with fractional equations.
Setting up the equation with fractions: y = 2/3x - 4 and y = 1/4x - 3/2.
Elimination of fractions by finding the least common multiple (LCM) of denominators.
Multiplying through by 12 to clear fractions and simplify the equation.
Solving for x after simplifying the equation to -5x = -30.
Finding x = 6 by dividing both sides by -5.
Substituting x = 6 into the original equation to solve for y.
Calculation of y = 0 using the value of x.
Presenting the solution as an ordered pair (x, y) = (6, 0).
Introduction to a problem with three variables and equations.
Setting up the system of equations with x, y, and z: x + y + z = 6, z = x + 2, y = 2x.
Substituting z and y in the first equation to solve for x.
Finding x = 1 by solving the equation 4x + 2 = 6.
Substituting x = 1 to find y and z.
Calculation of y = 2 and z = 3 using x = 1.
Presenting the solution as an ordered triple (x, y, z) = (1, 2, 3).
Transcripts
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