Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 10: Problem 11
Solve each nonlinear system of equations for real solutions. $$ \left\\{\begin{aligned} y &=x^{2}-3 \\ 4 x-y &=6 \end{aligned}\right. $$
Short Answer
Expert verified
The solutions are (3, 6) and (1, -2).
Step by step solution
01
Rewrite Second Equation
Start with the second equation of the system. We have: \[4x - y = 6\]Rearrange this to express \(y\) in terms of \(x\): \[y = 4x - 6\]
02
Substitute for y
Substitute the expression for \(y\) from the second equation into the first equation. We have: \[y = x^2 - 3\].Plug \(y = 4x - 6\) into the first equation:\[4x - 6 = x^2 - 3\]
03
Combine Like Terms
Solve the equation for \(x\):\[x^2 - 4x + 3 = 0\]This is a quadratic equation in standard form.
04
Solve the Quadratic Equation
Solve the quadratic equation using factorization. The given equation can be factored as:\(x^2 - 4x + 3 = (x-3)(x-1) = 0\)Thus, the solutions for \(x\) are \(x = 3\) and \(x = 1\).
05
Find Corresponding y-values
Substitute each \(x\) value back into the expression \(y = 4x - 6\) to find the corresponding \(y\) values:- If \(x = 3\), then: \[y = 4(3) - 6 = 12 - 6 = 6\]- If \(x = 1\), then: \[y = 4(1) - 6 = 4 - 6 = -2\]
06
Solution Conclusion
The solutions to the system of equations are the pairs \((x, y) = (3, 6)\) and \((x, y) = (1, -2)\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
Quadratic equations form a fundamental part of mathematics, expressing relationships in a square power of a variable. Generally, a quadratic equation appears as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.In the context of solving nonlinear systems, understanding quadratic equations is vital. It provides the structure to find unknown variables. A nonlinear system might combine a quadratic with a linear equation, as shown in the original exercise.To solve these equations:
- Identify the equation in the quadratic form.
- Use techniques like completing the square or factoring to find solutions.
- Recognize that there can be zero, one, or two real solutions depending on the discriminant \( b^2 - 4ac \).
Substitution Method
The substitution method is a strategic way to solve systems of equations, particularly useful when dealing with a mix of linear and quadratic equations. It involves replacing one variable in an equation with an equivalent expression from another.When using substitution:
- Start by solving one of the equations for one variable. For instance, solve for \( y = 4x - 6 \) in the second equation given in the problem.
- Substitute this expression into the other equation. In this case, replace \( y \) in \( y = x^2 - 3 \) with \( 4x - 6 \).
- This reduces the system into a single equation with one variable, which you can solve using appropriate algebraic methods.
Factoring Quadratics
Factoring is a powerful tool for solving quadratic equations. It allows you to break down a quadratic expression into products of simpler binomials.Here鈥檚 how to factor quadratic equations effectively:
- Firstly, rearrange the quadratic into standard form: \( ax^2 + bx + c = 0 \).
- Look for two numbers that multiply to \( ac \) (product) and add to \( b \) (sum).
- Use these numbers to split the middle term and factor by grouping if necessary.
- For the equation \( x^2 - 4x + 3 = 0 \), you recognize that \((x-3)(x-1) = 0\) are the factors, leading to solutions \( x = 3 \) and \( x = 1 \).
Solution Pairs
The concept of solution pairs is key in solving systems of equations. Each solution represents a point where all equations in the system meet or intersect. In other words, the solution pair \( (x, y) \) satisfies both equations simultaneously.To find solution pairs:
- Solve for one variable in terms of the other using techniques like substitution.
- Calculate corresponding values for the second variable for each solution of the first variable. For instance, substituting \( x = 3 \) or \( x = 1 \) into \( y = 4x - 6 \) gives \( y = 6 \) or \( y = -2 \) respectively.
- These correspond to the solution pairs \((3, 6)\) and \((1, -2)\).
