Problem 10 - Algebra and Trigonometry

Question

\(9-14\) . Use the elimination method to find all solutions of the system of equations. $$ \left\\{\begin{array}{l}{3 x^{2}+4 y=17} \\ {2 x^{2}+5 y=2}\end{array}\right. $$

Step-by-Step Solution

Answer

Solutions are \((x, y) = (\sqrt{11}, -4)\) and \((x, y) = (-\sqrt{11}, -4)\).

Step 1: Label the Equations

Let's label the given system of equations for convenience:1. \( 3x^2 + 4y = 17 \)2. \( 2x^2 + 5y = 2 \)

Step 2: Eliminate "y" by Equalizing Coefficients

To eliminate \(y\), we need the coefficients of \(y\) in both equations to be equal. Multiply the first equation by 5 and the second equation by 4:- Equation 1 scaled: \( 5(3x^2 + 4y) = 5(17) \) which simplifies to \( 15x^2 + 20y = 85 \)- Equation 2 scaled: \( 4(2x^2 + 5y) = 4(2) \) which simplifies to \( 8x^2 + 20y = 8 \)

Step 3: Subtract the Equations to Eliminate "y"

Subtract the scaled Equation 2 from the scaled Equation 1:\[(15x^2 + 20y) - (8x^2 + 20y) = 85 - 8\]This simplifies to:\[15x^2 - 8x^2 = 77\]Therefore, \(7x^2 = 77\).

Step 4: Solve for "x"

Divide both sides of the equation \(7x^2 = 77\) by 7:\[x^2 = \rac{77}{7} = 11\]Take the square root of both sides to solve for \(x\):\[x = \pm \sqrt{11}\]

Step 5: Substitute "x" Back to Solve for "y"

Substitute \(x = \sqrt{11}\) and \(x = -\sqrt{11}\) back into one of the original equations to find \(y\). Let's use Equation 1: \(3x^2 + 4y = 17\).- If \(x = \sqrt{11}\): \[ 3(11) + 4y = 17 \Rightarrow 33 + 4y = 17 \] Subtract 33 from both sides: \[ 4y = 17 - 33 = -16 \] \[ y = \frac{-16}{4} = -4 \]- If \(x = -\sqrt{11}\), the computation is the same since \(x^2\) doesn't change: \[ y = -4 \]

Step 6: Verify Solutions

Verify the solutions \((x, y) = (\sqrt{11}, -4)\) and \((x, y) = (-\sqrt{11}, -4)\) satisfy both equations:For Equation 1:- \( 3(11) + 4(-4) = 33 - 16 = 17 \Rightarrow \text{True} \)For Equation 2:- \( 2(11) + 5(-4) = 22 - 20 = 2 \Rightarrow \text{True} \)Both solutions satisfy the system.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with a common set of unknowns. In our exercise, we have a system consisting of two equations, each involving variables, in this case, \(x^2\) and \(y\). The primary goal when dealing with systems is to find values for the variables that satisfy all equations simultaneously.

• The given equations are quadratic in nature because they involve \(x^2\), a squared term. This means these are not simple linear equations.
• There are multiple methods to solve systems of equations, such as substitution, graphing, and the elimination method. In this case, the elimination method is used.
• The elimination method involves adding or subtracting equations to eliminate one of the variables, making it easier to solve for the remaining variable.

Understanding the relationships between the equations is crucial, as it helps in applying the right techniques to find the solution. Using these techniques, students can solve for one variable at a time.

Quadratic Equations

Quadratic equations are fundamental in algebra and appear when working with problems involving squared terms. These are expressions in which the variable is raised to the power of two, denoted as \(x^2\). The standard form of a quadratic equation is \(ax^2 + bx + c = 0\). However, when solving systems of equations, the quadratic terms might not always be isolated.

• In our exercise, both equations include quadratic terms \(3x^2\) and \(2x^2\).
• The elimination method is particularly useful when the quadratic terms across multiple equations allow for simplifying processes through algebraic manipulation.
• Once we aligned the equations to equate the coefficients of \(y\), subtraction eliminated the term, leaving a simple quadratic equation in terms of \(x\).

Recognizing and handling quadratic terms in the equations are key skills necessary for efficient problem-solving. Students should become comfortable with these as they frequently encounter them in diverse mathematical problems.

Algebraic Solution

Finding an algebraic solution involves systematically applying mathematical rules and operations to derive a solution. In algebra, solutions are typically found by isolating variables to find their value. Let's break down the steps involved:

• First, we labeled the equations for clarity, simplifying future steps.
• Multiplying each equation by coefficients allowed for eliminating \(y\). This leads to a simpler equation depending only on \(x^2\).
• Reducing the expression to \(7x^2 = 77\), we divided both sides by 7, resulting in \(x^2 = 11\).
• Solving \(x^2 = 11\) necessitates taking the square root, yielding \(x = \pm \sqrt{11}\), highlighting the potential for both positive and negative roots.
• With \(x\) values found, each is substituted back into one of the original equations (often the simplest one) to solve for \(y\). This is important as it checks consistency across the system.

Algebraic solutions demand both logical reasoning and mathematical operations. This example illustrates core techniques that form the basis of much of algebra, solidifying one's capability to address similar challenges.