Problem 85 - Precalculus

Question

The system of equations $$ \begin{aligned} x^{2}+2 y^{2} &=11 \\ 4 x^{2}+y^{2} &=16 \end{aligned} $$ can be solved by a change of variables. Taking \(u=x^{2}\) and \(v=y^{2},\) we can transform the system into $$ \begin{aligned} u+2 v &=11 \\ 4 u+v &=16 \end{aligned} $$ Find the solutions of the original system.

Step-by-Step Solution

Answer

The solutions are \((\sqrt{3}, 2), (\sqrt{3}, -2), (-\sqrt{3}, 2), (-\sqrt{3}, -2)\).

Step 1: Substitute variables

Let \( u = x^2 \) and \( v = y^2 \). Substituting into the original system, we have two equations: \( u + 2v = 11 \) and \( 4u + v = 16 \).

Step 2: Solve the linear system

First, solve the system of equations \( u + 2v = 11 \) and \( 4u + v = 16 \).Multiply the first equation by 4 to eliminate \( u \):\[ 4u + 8v = 44 \]Subtract the second equation from the newly formed equation:\[ (4u + 8v) - (4u + v) = 44 - 16 \]This simplifies to:\[ 7v = 28 \]\[ v = 4 \]

Step 3: Substitute the value of v

Substitute \( v = 4 \) back into \( u + 2v = 11 \):\[ u + 2(4) = 11 \]\[ u + 8 = 11 \]\[ u = 3 \]

Step 4: Revert to original variables

Recall \( u = x^2 \) and \( v = y^2 \). Thus, \( x^2 = 3 \) and \( y^2 = 4 \).

Step 5: Solve for x and y

Solving \( x^2 = 3 \) gives \( x = \pm \sqrt{3} \). Solving \( y^2 = 4 \) gives \( y = \pm 2 \).

Step 6: Determine all solutions

Combine the solutions for \( x \) and \( y \). The full list of solutions is:1. \( (x,y) = (\sqrt{3}, 2) \)2. \( (x,y) = (\sqrt{3}, -2) \)3. \( (x,y) = (-\sqrt{3}, 2) \)4. \( (x,y) = (-\sqrt{3}, -2) \)

Key Concepts

Change of Variables

In solving complex systems of equations, sometimes it helps to make a change of variables. This method simplifies the system by transforming variables into new ones. For instance, if you have variables squared in your equations, it might be beneficial to switch to these squares as new variables.

This technique was used with the original system of equations. We set new variables as follows: let \( u = x^2 \) and \( v = y^2 \). By doing this, the quadratic system was gently transformed into a simpler linear system, which is much easier to tackle. This simplification often reveals underlying patterns or solutions that are not immediately apparent in the original form.

Linear Equations

Linear equations are those where variables are only to the first power and are not multiplied together. When equations are linear, they form straight lines on a graph, making them straightforward to solve.

In our new system, by changing the variables, we obtained two linear equations: \( u + 2v = 11 \) and \( 4u + v = 16 \). These equations are simpler because they do not involve powers greater than one, and we can apply a variety of straightforward methods, such as elimination or substitution, to find solutions. This is a major advantage of transforming a more complex system into a linear form.

Quadratic Equations

Quadratic equations are those where variables are raised to the second power. In the context of a system of equations, they can make the problem more challenging to solve directly. However, they often have interesting properties, such as symmetry, which makes them a key topic of study in algebra.

In the original system of equations, \( x^2 + 2y^2 = 11 \) and \( 4x^2 + y^2 = 16 \), the presence of variables squared indicates that we are dealing with quadratic equations. By converting these into linear equations through our change of variables, we could solve the quadratic nature indirectly. After solving the linear system, we reverted to the original variables to address the quadratic solutions.

Solution of Systems

The solution of a system of equations refers to the values of variables that satisfy all the equations simultaneously. Solving systems like the one in this exercise often involves finding common values that make all equations true.

In this exercise, the system was initially complex due to quadratic terms, but through clever substitution, it was transformed into a system of linear equations. By solving the linear system, we found \( u = 3 \) and \( v = 4 \). Reverting back to original variables gave us \( x^2 = 3 \) and \( y^2 = 4 \). Solving for \( x \) and \( y \), this yields four potential solutions, specifically pairs of values that meet the original equations: \( (\sqrt{3}, 2) \), \( (\sqrt{3}, -2) \), \( (-\sqrt{3}, 2) \), and \( (-\sqrt{3}, -2) \). Each pair fits the condition set by the original system.