Problem 12 - Precalculus: Mathematical for Calculus

Question

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 \((\sqrt{11}, -4)\) and \((-\sqrt{11}, -4)\).

Step 1: Align the Equations

First, write down the system of equations clearly: 1) \(3x^2 + 4y = 17\) 2) \(2x^2 + 5y = 2\). We will work to eliminate one of the variables.

Step 2: Eliminate One Variable

We can eliminate \(y\) by multiplying the first equation by 5 and the second equation by 4: - \(5(3x^2 + 4y) = 5 \times 17\) results in \(15x^2 + 20y = 85\). - \(4(2x^2 + 5y) = 4 \times 2\) results in \(8x^2 + 20y = 8\).

Step 3: Subtract the Equations

Subtract the second new equation from the first to eliminate \(y\):\((15x^2 + 20y) - (8x^2 + 20y) = 85 - 8\). This simplifies to \(7x^2 = 77\).

Step 4: Solve for \(x^2\)

Solve \(7x^2 = 77\) by dividing both sides by 7: \(x^2 = \frac{77}{7} = 11\). So, \(x^2 = 11\).

Step 5: Calculate \(x\)

Since \(x^2 = 11\), we find \(x\) by taking the square root: \(x = \pm \sqrt{11}\).

Step 6: Solve for \(y\) Using \(x = \sqrt{11}\)

Substitute \(x = \sqrt{11}\) into the first original equation: \(3(\sqrt{11})^2 + 4y = 17\). This simplifies to \(33 + 4y = 17\). Solve for \(y\): \(4y = 17 - 33\), which is \(4y = -16\). Thus, \(y = -4\).

Step 7: Solve for \(y\) Using \(x = -\sqrt{11}\)

Substitute \(x = -\sqrt{11}\) into the first original equation again: \(3(-\sqrt{11})^2 + 4y = 17\), which also simplifies to \(33 + 4y = 17\). Solve for \(y\): \(4y = 17 - 33\), which is \(4y = -16\). Thus, \(y = -4\).

Step 8: Conclusion

The solutions to the system are \((x, y) = (\sqrt{11}, -4)\) and \((-\sqrt{11}, -4)\).

Key Concepts

System of Equations

A **system of equations** is a set of two or more equations that share the same variables. In the given exercise, the system consists of two equations with variables \(x^2\) and \(y\):

• \(3x^2 + 4y = 17\)
• \(2x^2 + 5y = 2\)

The objective is to find the values of \(x\) and \(y\) that satisfy both equations simultaneously. Solving a system of equations can be done using various methods such as substitution, elimination, or graphical interpretation. In this exercise, the elimination method is employed to simplify the process. Each equation provides a different perspective on how \(x\) and \(y\) relate to each other, and the solution must satisfy both equations.
The key challenge is to work through the equations to simplify them until the variables can be clearly identified.

Quadratic Equations

A **quadratic equation** is a second-degree polynomial equation. In this context, the variable \(x^2\) forms the quadratic part within the system of equations. Quadratic equations typically take the form \(ax^2 + bx + c = 0\). Here, however, it is integrated into a system where the equation involves both \(x^2\) and \(y\).
To solve quadratic equations, we often use methods like factoring, completing the square, or the quadratic formula. However, within this system, the quadratic terms are handled through elimination, which involves manipulating the equations to either add or subtract terms, effectively removing one variable, and thereby resolving the quadratic aspect directly.
The aim is to focus on the quadratic term, manage it within the context of a system, and distill it down until simple algebraic manipulation can yield solutions.

Solving Equations

**Solving equations** refers to finding the value or set of values that satisfy the equation(s). In this particular problem, the solution involves finding \(x\) and \(y\) that satisfy both given equations simultaneously.
This process is carried out by systematically applying algebraic operations to both sides of the equations, maintaining the balance, and simplifying them step by step. In elimination, relevant equations are modified to achieve cancellation of one variable, leading to a more straightforward equation, which can then be solved algebraically.
The solution involves substituting the found value(s) back into the original equations to ensure they hold true. This verifies the accuracy of the solution and is critical for confirming that you have indeed solved the system of equations correctly.

Substitution Method

The **substitution method** is one of the strategies to solve systems of equations, although not directly used in this exercise, offering useful insight. Instead of eliminating variables, substitution involves solving one of the equations for one variable and substituting that expression in the other equation.
This method helps in situations where one equation can be easily solved for a variable, making it straightforward to substitute back into other equations. While our current problem calls for elimination due to ease in manipulating coefficients, substitution can be empowering in other systems where solving for a variable first simplifies the entire problem.
It can provide clarity when one equation is too complex but can become simple with a substitution, highlighting its valued role in achieving solutions for any system of equations.