Problem 27 - Algebra and Trigonometry

Question

Solve each system by the addition method. \(\left\\{\begin{array}{l}{3 x-4 y=11} \\ {2 x+3 y=-4}\end{array}\right.\)

Step-by-Step Solution

Answer

The solutions of the system are points (-1, \sqrt{3}),(-1,-\sqrt{3}),(0,2), and (0,-2)

Step 1: Rearrange the equations

Rearrange the first equation \(y^{2}-x=4\) to \(x = y^{2} - 4\). The second equation remains the same: \(x^{2}+y^{2}=4\)

Step 2: Substitute x in the second equation

Substitute the value of \(x\) from the first equation into the second equation, giving \((y^{2} - 4)^{2} + y^{2} = 4\). This simplifies to \(y^{4} - 8y^{2} + 16 + y^{2} = 4\), which further simplifies to \(y^{4} - 7y^{2} + 12 = 0\)

Step 3: Solve the fourth-grade equation

Let \(m = y^{2}\). The equation becomes a quadratic equation: \(m^{2} - 7m + 12 = 0\). Solve this equation to find the roots, using the quadratic formula \(m = (7 \pm \sqrt{49 - 4*12}) / 2\). The roots of this equation are \(m = 3 and m = 4\). Here, \(m\) represents \(y^{2}\), so \(y^{2}\) is 3 and 4. Taking the square root of these values will give \(y\) = \( \pm \sqrt{3} \) and \(y\) = \( \pm 2\)

Step 4: Substitute y in the first equation

Substitute the obtained values of \(y\) back into the rearranged first equation to solve for \(x\). For \(y = \sqrt{3}\), \(x = (\sqrt{3})^{2} - 4 = -1\). For \(y = - \sqrt{3}\), \(x = (-\sqrt{3})^{2} - 4 = -1\). For \(y = 2\), \(x = 2^{2} - 4 = 0\). For \(y = -2\), \(x = (-2)^{2} - 4 = 0\).

Key Concepts

Addition Method

The addition method is a technique used to solve a system of equations by eliminating one variable. This method works well when the coefficients of one of the variables in the different equations are additive inverses. It requires careful rearrangement and addition of the equations. In some cases, equations might need to be multiplied by constants to align the terms for cancellation.
Through addition, one variable is removed, and you're left with a simpler equation. This makes it easier to solve for the remaining unknowns.
To implement the addition method effectively, follow these steps:

• First, identify a variable to eliminate by aligning the equations.
• Make the coefficient of that variable opposite in the equations, if necessary.
• Add the equations, which should result in an equation with one less variable.

This method leverages the principle that adding equal quantities to both sides of an equation keeps the solution intact.

Quadratic Equation

Quadratic equations take the form of a second-degree polynomial equation: \(ax^2 + bx + c = 0\). In our system, the substitution method introduced a new unknown that led to a quadratic equation.
Quadratic equations are common in many algebraic scenarios and are especially important because they can model real-world problems where there is curvature in the relationships between variables.
Solutions can be found using:

• Factoring, if the equation can be rewritten as a product of binomials.
• Completing the square, which makes the quadratic a perfect square trinomial.
• The quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Understanding the nature of quadratic equations is crucial for building a robust foundation in algebra. They frequently appear in conjunction with other algebraic solving techniques.

Substitution Method

The substitution method involves solving one of the equations in a system for one variable and then substituting that expression into the other equation. This allows you to work with a single equation in terms of one variable, making it simpler to solve.
To use this method effectively:

• Rearrange one of the equations to express one variable in terms of the other variables.
• Substitute that expression into the other equation(s) to eliminate one variable.
• Solve the resulting equation(s) for the unknown(s).

In the provided problem, the first equation was rearranged to express \(x\) in terms of \(y\). This expression was then substituted into the second equation, which facilitated the solution.
Substitution is a versatile method, effective for linear and non-linear systems alike.

Algebraic Solution

An algebraic solution involves solving equations using algebraic manipulations and techniques, adhering to principles of algebra to obtain solutions. This approach involves defining the variables, setting up equations, and simplifying them systematically until you find the values of the unknowns.
The process typically consists of:

• Carefully examining the given system of equations.
• Rearranging and simplifying equations as needed, using methods like addition, substitution, and elimination.
• Solving for each variable systematically.

The aim is to find values for unknowns that satisfy all given equations in the system.
An algebraic solution provides a rigorous and consistent approach to solving both simple and complex systems of equations, forming the backbone of problem-solving in algebra.