Problem 84 - Precalculus

Question

The system of equations $$ \begin{array}{l} x^{2}+y^{2}=4 \\ x^{2}-y^{2}=2 \end{array} $$ can be solved by a change of variables. Taking \(u=x^{2}\) and \(v=y^{2},\) we can transform the system into $$ \begin{array}{l} u+v=4 \\ u-v=2 \end{array} $$ Find the solutions of the original system.

Step-by-Step Solution

Answer

Solutions are \((\sqrt{3}, 1)\), \((\sqrt{3}, -1)\), \((-\sqrt{3}, 1)\), \((-\sqrt{3}, -1)\)."

Step 1: Change Variables

We start by making the substitutions \( u = x^2 \) and \( v = y^2 \). This transforms the original system of equations into: \( u + v = 4 \) and \( u - v = 2 \).

Step 2: Solve the New System

We now solve the system of equations: 1. \( u + v = 4 \) 2. \( u - v = 2 \)Add equation (1) and (2) together:\[ (u + v) + (u - v) = 4 + 2 \]\[ 2u = 6 \]\[ u = 3 \]

Step 3: Find v

Next, substitute \( u = 3 \) back into one of the equations, for example, \( u + v = 4 \):\[ 3 + v = 4 \]\[ v = 1 \]

Step 4: Find x and y

Now we substitute back to find \( x \) and \( y \) from \( u = x^2 \) and \( v = y^2 \):1. \( x^2 = 3 \)2. \( y^2 = 1 \)This gives us two pairs of solutions:\( x = \sqrt{3} \) or \( x = -\sqrt{3} \) and \( y = 1 \) or \( y = -1 \).

Step 5: Compile All Solutions

The pairs \( (x, y) \) that solve the original system are:1. \((\sqrt{3}, 1)\)2. \((\sqrt{3}, -1)\)3. \((-\sqrt{3}, 1)\)4. \((-\sqrt{3}, -1)\).

Key Concepts

Change of Variables

The change of variables is a strategy used to simplify complex equations or systems of equations. In essence, it involves substituting parts of an equation with variables that make the problem more manageable. This helps in reducing the complexity and often transforms it into a more familiar format.

In the given problem, we used the change of variables by letting \( u = x^2 \) and \( v = y^2 \). By doing this, the complex quadratic relationships among \(x\) and \(y\) become linear equations in terms of \(u\) and \(v\). The original system of equations, which was difficult to solve directly, was transformed into:

• \( u + v = 4 \)
• \( u - v = 2 \)

Converting a problem into a simpler form is a common mathematical strategy and the change of variables is one of the most powerful tools in the toolbox for doing this.

Algebraic Manipulation

Algebraic manipulation involves the use of algebraic techniques to simplify and solve equations. After substituting variables, the equations \( u + v = 4 \) and \( u - v = 2 \) become straightforward to solve.

The next step is to add these two equations together. This is a common technique called elimination, which helps in removing one of the variables. By adding \( u + v \) and \( u - v \), the \( v \) cancels out:

• \((u + v) + (u - v) = 4 + 2\)
• This results in \(2u = 6\)
• Hence, \(u = 3\)

Once we have \( u \), we can substitute back into one of the original equations to find \( v \). By substituting \( u = 3 \) into \( u + v = 4 \):

• \(3 + v = 4\)
• Thus, \(v = 1\)

Using algebraic manipulation, we can systematically solve for the unknowns in a very methodical way.

Solving Quadratic Equations

Once the values of \( u \) and \( v \) are identified, the final step involves finding \( x \) and \( y \) using the defined relation between these variables. In this problem, \( u = x^2 \) and \( v = y^2 \).

Now, with \( u = 3 \) and \( v = 1 \), you solve each quadratic equation:

• \( x^2 = 3 \) implies \( x = \pm \sqrt{3} \)
• \( y^2 = 1 \) implies \( y = \pm 1 \)

Quadratic equations can have more than one solution because they involve squared terms. Hence, for each equation, we consider both the positive and negative roots. This means that there are multiple sets of \( (x, y) \) pairs that satisfy the original system:

• \( (\sqrt{3}, 1) \)
• \( (\sqrt{3}, -1) \)
• \( (-\sqrt{3}, 1) \)
• \( (-\sqrt{3}, -1) \)

Understanding how to solve quadratic equations is crucial because they often appear in many real-world problems and theoretical scenarios.