Problem 31
Question
In Exercises \(29-42,\) solve each system by the method of your choice. $$ \left\\{\begin{array}{l} 2 x^{2}+y^{2}=18 \\ x y=4 \end{array}\right. $$
Step-by-Step Solution
Verified Answer
The solution set for the system of equations is \{ (\sqrt{2}, 2\sqrt{2}), (-\sqrt{2}, -2\sqrt{2}), (2\sqrt{2}, \sqrt{2}), (-2\sqrt{2}, -\sqrt{2})\}.
1Step 1: Express y in terms of x from the second equation
The second equation is \( xy = 4 \). We can solve for y by dividing both sides by x to get \( y = 4/x \).
2Step 2: Substitute \( y = 4/x \) into the first equation
By substituting \( y = 4/x \) into the first equation \(2x^2 + y^2 = 18\), we get \(2x^2 + (4/x)^2 = 18\). Multiply every term by \(x^2\) to eliminate the denominator in the second term. This transforms the equation to a quadratic equation \(2x^4 + 16 = 18x^2\).
3Step 3: Rewrite the equation and solve for x
Rewrite the equation from Step 2 as \(2x^4 - 18x^2 + 16 = 0\), we can now solve this as a quadratic equation with \(a = 2, b = -18, c = 16\). This yields \(x^2 = 2, x^2 = 8\). Therefore, \(x = \sqrt{2}, -\sqrt{2}, 2\sqrt{2}, -2\sqrt{2}\).
4Step 4: Solve for y using the value of x
Substitute each of the obtained x-values from Step 3 into the equation \( y = 4/x \) to get the corresponding y-values. This gives \( y = 2\sqrt{2}, -2\sqrt{2}, \sqrt{2}, -\sqrt{2} \). Therefore, the solution set for the system of equations is \{ (\sqrt{2}, 2\sqrt{2}), (-\sqrt{2}, -2\sqrt{2}), (2\sqrt{2}, \sqrt{2}), (-2\sqrt{2}, -\sqrt{2})\}.
Key Concepts
Quadratic EquationsSubstitution MethodAlgebraic Expressions
Quadratic Equations
A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). It is called quadratic because "quad" refers to a square, indicating the highest degree term \( x^2 \). Quadratic equations can be solved by various methods such as factoring, using the quadratic formula, or completing the square.
In this exercise, the first equation after substitution transforms into \( 2x^4 - 18x^2 + 16 = 0 \). Though it doesn't look like a traditional quadratic equation at first glance, it can be treated as one by considering \( x^2 \) as a temporary variable.
This approach simplifies solving by letting us apply quadratic methods. Solving the resulting equation yields the possible values for \( x^2 \), which we then use to find the specific solutions for \( x \). These solutions can then be explored further to find corresponding \( y \)-values in the system of equations.
In this exercise, the first equation after substitution transforms into \( 2x^4 - 18x^2 + 16 = 0 \). Though it doesn't look like a traditional quadratic equation at first glance, it can be treated as one by considering \( x^2 \) as a temporary variable.
This approach simplifies solving by letting us apply quadratic methods. Solving the resulting equation yields the possible values for \( x^2 \), which we then use to find the specific solutions for \( x \). These solutions can then be explored further to find corresponding \( y \)-values in the system of equations.
Substitution Method
The substitution method is a popular technique for solving systems of equations, especially when dealing with non-linear equations like quadratic equations. The key idea is to solve one equation for one variable and substitute that expression into the other equation. This reduces the system of equations to a single equation with one unknown.
In this particular system, the substitution method begins by expressing \( y \) from the second equation \( xy = 4 \). This is straightforward, yielding \( y = \frac{4}{x} \).
In this particular system, the substitution method begins by expressing \( y \) from the second equation \( xy = 4 \). This is straightforward, yielding \( y = \frac{4}{x} \).
- First, solve for one variable in one of the equations.
- Second, substitute this expression into the other equation. Here, \( y = 4/x \) is substituted into \( 2x^2 + y^2 = 18 \).
- Finally, solve the resulting equation, which in this case becomes a quadratic equation.
Algebraic Expressions
Algebraic expressions are combinations of numbers, variables, and mathematical operations like addition, multiplication, and exponentiation. Understanding how to manipulate these expressions is fundamental in solving equations.
For example, in our exercise, we began by solving for \( y \) as \( 4/x \). Then, we substituted this expression into the quadratic equation, resulting in \( 2x^4 + 16 = 18x^2 \).
For example, in our exercise, we began by solving for \( y \) as \( 4/x \). Then, we substituted this expression into the quadratic equation, resulting in \( 2x^4 + 16 = 18x^2 \).
- Algebraically, we simplified this to \( 2x^4 - 18x^2 + 16 = 0 \), using operations like subtraction and multiplication to eliminate fractions.
- This simplifies the complexity of the original system and allows for the application of quadratic solution strategies.
Other exercises in this chapter
Problem 31
write the partial fraction decomposition of each rational expression. $$ \frac{5 x^{2}+6 x+3}{(x+1)\left(x^{2}+2 x+2\right)} $$
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You throw a ball straight up from a rooftop. The ball misses the rooftop on its way down and eventually strikes the ground. A mathematical model can be used to
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In Exercises \(31-42,\) solve by the method of your choice. Identify systems with no solution and systems with infinitely many solutions, using set notation to
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