Problem 13 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: Express both equations

Given the system of equations: \[ 3x^2 - y^2 = 11 \] and \[ x^2 + 4y^2 = 8 \] We need to solve these using elimination.

Step 2: Align equations for elimination

Notice that the second equation is \( x^2 + 4y^2 = 8 \). We can manipulate the first equation to make the structure similar by solving for one variable.

Step 3: Manipulate the first equation

Rewrite the first equation: \[ 3x^2 - y^2 = 11 \] to possibly make the coefficients of \(y^2\) in both equations similar. Start by multiplying the entire second equation by 1, so we can add or subtract without fractions.

Step 4: Eliminate the y-term

Now multiply the second equation by 3: \[ 3(x^2 + 4y^2) = 3(8) \] resulting in \[ 3x^2 + 12y^2 = 24 \]. Subtract the first equation from this result: \[ (3x^2 + 12y^2) - (3x^2 - y^2) = 24 - 11 \].

Step 5: Simplify resulting equation

After subtraction, simplify the equation: \[ 13y^2 = 13 \], which gives \( y^2 = 1 \).

Step 6: Solve for y

From \( y^2 = 1 \), we have \( y = 1 \) or \( y = -1 \).

Step 7: Substitute y-values into one of the original equations

Substitute \( y = 1 \) into \( x^2 + 4y^2 = 8 \): \[ x^2 + 4(1)^2 = 8 \]. Simplifying gives \( x^2 + 4 = 8 \), hence \( x^2 = 4 \).

Step 8: Solve for x when y = 1

This results in \( x = 2 \) or \( x = -2 \). Thus, the solutions when \( y = 1 \) are \((2, 1)\) and \((-2, 1)\).

Step 9: Repeat substitution for y = -1

Substitute \( y = -1 \) into \( x^2 + 4y^2 = 8 \): \[ x^2 + 4(-1)^2 = 8 \]. This simplifies to \( x^2 = 4 \).

Step 10: Solve for x when y = -1

Similar to the previous step, \( x = 2 \) or \( x = -2 \). Thus, the solutions when \( y = -1 \) are \((2, -1)\) and \((-2, -1)\).

Key Concepts

Understanding Systems of Equations

Systems of equations are a set of two or more equations with the same variables. Solving systems of equations means finding the values for the variables that satisfy all equations simultaneously.In our exercise, we are given two equations involving the variables x and y:

• The first equation is a quadratic equation of the form: \(3x^2 - y^2 = 11\)
• The second equation, also quadratic, is: \(x^2 + 4y^2 = 8\)

These equations form a system because they involve the same variables, x and y, and we must find a common solution that satisfies both equations.The elimination method is used here to remove one of the variables by adding or subtracting equations. This simplifies the system, making it easier to solve.

Exploring Quadratic Equations

Quadratic equations are polynomial equations of degree two. They typically have the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants.In our given system, each equation is quadratic, as they involve terms like \(x^2\) and \(y^2\) but do not necessarily include linear terms, making some solutions unique.Quadratic equations can have two, one, or no real solutions, depending on the discriminant value derived from the coefficients.In our exercise, we identify that each equation can potentially have two solutions when solved independently for a single variable. However, since these equations are part of the system, the overall solution must satisfy both, leading to specific points of intersection when plotted on a graph as curves.Solutions for quadratic equations in the system are pinpointed by substituting values between equations, which can reveal points like \(x = 2\), \(y = 1\), \(-2, 1\), etc.

The Power of Algebraic Manipulation

Algebraic manipulation is an essential tool in solving equations and functions efficiently.It involves rearranging and simplifying expressions and equations to make them solvable; like in our exercise where the elimination method was applied.By understanding algebraic properties:

• You can add or subtract equations to eliminate variables.
• Multiply both sides of an equation by the same number to align terms for elimination.

In this exercise, to eliminate \(y^2\), the second equation was multiplied by 3, then subtracted from the first. This eliminated \(x^2\) and \(y^2\) terms, making it simpler to solve \(13y^2 = 13\).Such manipulations help solve complex equations where straightforward substitution is not feasible, especially in non-linear systems like those involving quadratics.By mastering these techniques, students can tackle a wide range of mathematical problems effectively.