Problem 37
Question
Solve each system by the method of your choice. $$\left\\{\begin{array}{l} x^{2}+(y-2)^{2}-4 \\ x^{2}-2 y-0 \end{array}\right.$$
Step-by-Step Solution
Verified Answer
The solutions for the given system of equations are (0,0) and (\(±\sqrt{2}\),\(\frac{8}{5}\)).
1Step 1: Analysing the Given Equations
The two given equations are: \[x^{2}+(y-2)^{2}=4\] and \[x^{2}=2y\]. This is a system of non-linear equations. Notice that both equations contain \(x^{2}\), thus, it would be useful to use substitution method.
2Step 2: Substitution
Taking \(x^{2} =2y\) from the second equation and placing it into the first equation, we get: \[(2y)+(y-2)^{2}=4\]. Simplifying, we get: \[5y^{2}-8y=0\].
3Step 3: Solving for y
The quadratic equation can be further simplified by factorising: \[y(5y-8)=0\]. Thus, the possible values of y are : 0 and \(\frac{8}{5}\).
4Step 4: Solving for x
Substituting y values into the second equation \(x^{2}=2y\) , we get x=0, when y=0 and x=±2 when y=\(\frac{8}{5}\).
Key Concepts
Substitution MethodQuadratic EquationsFactorizationAlgebraic Solutions
Substitution Method
The substitution method is one of the essential techniques for solving systems of equations. In this method, you solve one of the equations for a single variable and substitute that solution into the other equation. This approach simplifies the equations, allowing you to find the solution step-by-step more efficiently.
In our original exercise, the second equation, \(x^2 = 2y\), is already solved for \(x^2\). By inserting \(x^2 = 2y\) into the first equation \(x^2 + (y-2)^2 = 4\), you can reduce the complexity of the system. This substitution helps transform a potentially intimidating system into a more manageable form, making it easier to proceed to the next steps in finding a solution.
In our original exercise, the second equation, \(x^2 = 2y\), is already solved for \(x^2\). By inserting \(x^2 = 2y\) into the first equation \(x^2 + (y-2)^2 = 4\), you can reduce the complexity of the system. This substitution helps transform a potentially intimidating system into a more manageable form, making it easier to proceed to the next steps in finding a solution.
Quadratic Equations
Quadratic equations play a crucial role in algebra and can appear intimidating. However, they often have straightforward solutions. A typical quadratic equation takes the form \(ax^2 + bx + c = 0\). The solution typically revolves around finding the values of the variable that make the equation true.
In our example, once substitution is made, the resulting equation becomes \(5y^2 - 8y = 0\). Although this seems different at first, it is indeed a quadratic equation with \(a = 5\), \(b = -8\), and \(c = 0\). Recognizing the structure allows us to apply standard techniques like factorization to find the solutions.
In our example, once substitution is made, the resulting equation becomes \(5y^2 - 8y = 0\). Although this seems different at first, it is indeed a quadratic equation with \(a = 5\), \(b = -8\), and \(c = 0\). Recognizing the structure allows us to apply standard techniques like factorization to find the solutions.
Factorization
Factorization is a powerful algebraic technique used to simplify expressions and solve equations. The idea is to express a polynomial as a product of its factors. This method can make it easier to find solutions to equations, especially quadratics.
For our quadratic equation \(5y^2 - 8y = 0\), factorization can be applied directly. We can factor out the common term \(y\), which gives us \(y(5y - 8) = 0\). This reveals the possible solutions more clearly because each factor corresponds to a potential solution.
For our quadratic equation \(5y^2 - 8y = 0\), factorization can be applied directly. We can factor out the common term \(y\), which gives us \(y(5y - 8) = 0\). This reveals the possible solutions more clearly because each factor corresponds to a potential solution.
- Set each factor equal to zero: \(y = 0\) or \(5y - 8 = 0\).
- Solving these gives \(y = 0\) and \(y = \frac{8}{5}\).
Algebraic Solutions
Algebraic solutions involve finding the values of variables that satisfy the given conditions of an equation or system of equations. In this context, nonlinear systems present unique challenges, but the techniques we've discussed lead to clear solutions.
After determining possible values for \(y\) using factorization, we return to find corresponding values for \(x\) using algebraic manipulation and substitution. By substituting \(y = 0\) into \(x^2 = 2y\), we get \(x^2 = 0\), leading to \(x = 0\). Similarly, for \(y = \frac{8}{5}\), substituting into the same equation gives \(x^2 = 2 \times \frac{8}{5} = \frac{16}{5}\), resulting in \(x = \pm \sqrt{\frac{16}{5}}\) which simplifies to \(x = \pm 2\).
After determining possible values for \(y\) using factorization, we return to find corresponding values for \(x\) using algebraic manipulation and substitution. By substituting \(y = 0\) into \(x^2 = 2y\), we get \(x^2 = 0\), leading to \(x = 0\). Similarly, for \(y = \frac{8}{5}\), substituting into the same equation gives \(x^2 = 2 \times \frac{8}{5} = \frac{16}{5}\), resulting in \(x = \pm \sqrt{\frac{16}{5}}\) which simplifies to \(x = \pm 2\).
- Thus, the solutions are: when \(y = 0\), \(x = 0\); when \(y = \frac{8}{5}\), \(x = \pm 2\).
- These solutions are the critical final step in validating the correct values that satisfy the initial system of equations.
Other exercises in this chapter
Problem 37
Write the partial fraction decomposition of each rational expression. $$\frac{x^{3}+x^{2}+2}{\left(x^{2}+2\right)^{2}}$$
View solution Problem 37
In Exercises 27–62, graph the solution set of each system of inequalities or indicate that the system has no solution. $$-2 \leq x
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Solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to express their solution s
View solution Problem 38
Exercises 37-39 will help you prepare for the material covered in the first section of the next chapter. Solve the system: $$\left\\{\begin{aligned}w-x+2 y-2 z
View solution