Linear and quadratic systems — Basic example | Math | SAT | Khan Academy
Summary
TLDRThe teacher guides students through solving a system of equations, where one equation is linear and the other is nonlinear. By using substitution, the teacher demonstrates how to solve for the variable Y, leading to a quadratic equation that can be factored. The solution yields Y-values of 5 and -3. The teacher then verifies these solutions by substituting them back into the original equations and matching them with the provided answer choices, emphasizing a strategic approach to solving under time pressure.
Takeaways
- 📚 The script discusses solving a system of equations where the first equation is linear and the second is nonlinear due to the presence of a squared term in Y.
- 🔍 The teacher suggests using substitution to solve the system, starting with the linear equation to express X in terms of Y.
- 🔄 The substitution involves replacing X in the nonlinear equation with the expression derived from the linear equation, which is Y squared minus nine.
- 🧩 After substitution, the teacher rearranges the equation to isolate Y, resulting in a quadratic equation Y squared minus 2Y minus 15 equals zero.
- ➖ The teacher advises subtracting 2Y and six from both sides to simplify the equation and prepare it for factoring.
- 🔢 The quadratic is factored by finding two numbers that multiply to -15 and add up to -2, which are -5 and 3.
- 📝 The factored form of the equation is (Y - 5)(Y + 3) = 0, indicating that the solutions for Y are 5 and -3.
- 🎯 The teacher emphasizes that both Y = 5 and Y = -3 are solutions to the system, as they satisfy the quadratic equation.
- 🔄 To find the corresponding X values, the teacher suggests substituting the Y values back into the original equations.
- 📌 The script clarifies that the solutions are pairs of (X, Y), not just individual values, and that the correct solutions are when Y = 5, X = 16, and when Y = -3, X = 0.
- 🕵️♂️ The teacher also points out a common mistake where one might confuse the coordinates, emphasizing the importance of identifying the correct (X, Y) pairs.
- 🔄 The final step involves verifying the solutions by substituting the Y values back into the linear equation to ensure the X values are correct.
Q & A
What type of system of equations is discussed in the script?
-The script discusses a system of equations that includes a linear equation and a nonlinear one, specifically with a quadratic term in the second equation.
What is the first step suggested to solve the system of equations?
-The first step suggested is to use substitution, where the expression for X from the first equation is substituted into the second equation.
What is the expression for X in the first equation?
-The expression for X in the first equation is 2y + 6.
How is the second equation transformed after substituting the expression for X?
-After substituting, the second equation becomes a quadratic equation: y^2 - 2y - 15 = 0.
What method is used to solve the quadratic equation y^2 - 2y - 15 = 0?
-The method used to solve the quadratic equation is factoring.
What are the two numbers that factor the quadratic expression y^2 - 2y - 15?
-The two numbers that factor the quadratic expression are -5 and 3.
What are the solutions for Y from the factored equation?
-The solutions for Y are 5 and -3.
How can you find the corresponding X values for the given Y values?
-You can find the corresponding X values by substituting the Y values back into the first equation (2y + 6).
What are the X values when Y equals 5 and when Y equals -3?
-When Y equals 5, X equals 16 (2*5 + 6). When Y equals -3, X equals 0 (2*(-3) + 6).
Why might the script suggest reviewing factoring quadratic expressions on Khan Academy if you find it confusing?
-The script suggests reviewing on Khan Academy to gain a better understanding of the factoring process and to clarify any confusion about the method used to solve the quadratic equation.
What is the final step to ensure the correctness of the solutions found?
-The final step is to substitute the found Y values and their corresponding X values back into the original system of equations to verify that they satisfy both equations.
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