Problem 11 - Precalculus: Mathematics for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{aligned} 3 x^{2}-y^{2} &=11 \\ x^{2}+4 y^{2} &=8 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \((2, 1), (2, -1), (-2, 1), (-2, -1)\).

Step 1: Arrange the equations

Write the system of equations clearly for reference. We have:1. \(3x^2 - y^2 = 11\)2. \(x^2 + 4y^2 = 8\)

Step 2: Multiply to eliminate terms

Multiply the first equation by 4 and the second equation by 1 to align the coefficients of \(y^2\). The equations become:1. \(12x^2 - 4y^2 = 44\)2. \(x^2 + 4y^2 = 8\)

Step 3: Add the equations

Add the modified equations to eliminate \(y^2\):\[(12x^2 - 4y^2) + (x^2 + 4y^2) = 44 + 8\]This simplifies to:\[13x^2 = 52\]

Step 4: Solve for \(x\)

Divide both sides by 13 to isolate \(x^2\):\[x^2 = 4\]Taking the square root of both sides gives \(x = 2\) or \(x = -2\).

Step 5: Substitute \(x\) to find \(y\)

Substitute \(x = 2\) into one of the original equations, say the second:\[2^2 + 4y^2 = 8\]Simplifying, we find:\[4 + 4y^2 = 8 \4y^2 = 4 \y^2 = 1\]Hence, \(y = 1\) or \(y = -1\).Substituting \(x = -2\) into the same equation yields the same values for \(y\).

Step 6: Verify solutions

We have potential solutions: \((x, y) = (2, 1), (2, -1), (-2, 1), (-2, -1)\).Check each in the original equations to confirm they satisfy both:For example, substitute \((2, 1)\) into both equations:1. \(3(2)^2 - (1)^2 = 12 - 1 = 11\)\, correct.2. \((2)^2 + 4(1)^2 = 4 + 4 = 8\)\, correct.Repeat for \((2, -1), (-2, 1), (-2, -1)\).

Step 7: List all verified solutions

After verifying, the solutions to the system are:\((x, y) = (2, 1), (2, -1), (-2, 1), (-2, -1)\).

Key Concepts

systems of equations

A system of equations is essentially a set where two or more equations are solved together, usually to find several unknowns. In our exercise, the system consists of two equations involving variables \(x\) and \(y\). These systems can include several types of equations, such as linear or quadratic, and here we have one quadratic system. Solving systems of equations means finding values for the variables that make all the equations true simultaneously.
For our specific problem, the goal was to find values for \(x\) and \(y\) that satisfy both equations:

• \(3x^2 - y^2 = 11\)
• \(x^2 + 4y^2 = 8\)

In such systems, it's common to use methods like substitution or elimination. Both are geared toward reducing the number of variables and simplifying equations to reach a solution.
However, given the quadratic nature of our equations, the elimination method was chosen due to its ability to quickly eliminate one of the variables. This allows us to solve them sequentially.

solving quadratic equations

Quadratic equations, characterized by terms like \(ax^2 + bx + c = 0\), are a fundamental part of algebra. In our exercise, each equation can lead to a quadratic form when transformed properly. The standard procedure for solving involves methods like:

• Factoring
• Completing the square
• Quadratic formula

However, in this instance, once we used the elimination process, the resulting equation \(13x^2 = 52\) was already simple enough to solve directly. By dividing both sides by 13, we simplified it to \(x^2 = 4\).
From here, taking the square root of both sides was straightforward and yielded the solutions \(x = 2\) or \(x = -2\). These potential solutions were critically important, as they provided the groundwork for solving for \(y\) using substitution back into the original equations.

substitution

The substitution method involves replacing a variable in one equation with an expression derived from another. It's a key technique when dealing with systems of equations because it simplifies the equations step by step. In our task, once we found the possible values for \(x\) (i.e., \(x = 2\) and \(x = -2\)), substitution allowed us to solve for \(y\).
Taking one value of \(x\), such as \(x = 2\), and substituting it back into one of the original equations, specifically:

• \(x^2 + 4y^2 = 8\)

turns the equation to:

• \(4 + 4y^2 = 8\)

Solving this, we find \(y^2 = 1\), leading to two possible solutions for \(y\): \(y = 1\) or \(y = -1\).
This process was repeated for both values of \(x\), allowing us to identify all pairs of \((x, y)\) that satisfy both original equations. Such methodical substitution is crucial as it efficiently narrows down the potential solutions to the validated ones listed at the end of the exercise.