Problem 10 - Precalculus: Mathematics 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: Write the System of Equations

We start by writing the given system of equations clearly: 1. \(3x^2 + 4y = 17\)2. \(2x^2 + 5y = 2\)

Step 2: Eliminate y by Equalizing Coefficients

To eliminate \(y\), we align the coefficients of \(y\) in both equations. Multiply the first equation by 5 and the second by 4 to equalize the \(y\) coefficients.Equation 1: \[5(3x^2 + 4y) = 5(17)\] gives: \(15x^2 + 20y = 85\).Equation 2: \[4(2x^2 + 5y) = 4(2)\] gives: \(8x^2 + 20y = 8\).

Step 3: Subtract the New Equations

Subtract the second 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

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

Step 5: Substitute Back to Find y

Substitute \(x = \sqrt{11}\) and \(x = -\sqrt{11}\) back into one of the original equations (e.g., \(3x^2 + 4y = 17\)) to find \(y\). For \(x = \sqrt{11}\):\[3(11) + 4y = 17\]\[33 + 4y = 17\]\[4y = 17 - 33\]\[4y = -16\]\[y = -4\].For \(x = -\sqrt{11}\), the process is identical, leading to the same value for \(y: y = -4\).

Step 6: Conclusion

The solutions for the system are \((\sqrt{11}, -4)\) and \((-\sqrt{11}, -4)\).

Key Concepts

Systems of Equations

When dealing with systems of equations, the goal is to find a set of values for the variables that satisfy all equations in the system simultaneously. In our exercise, the system consists of two equations:

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

This system is called a non-linear system because it contains quadratic terms, specifically \(x^2\). Solving these types of systems often requires the user to employ specific techniques such as substitution or elimination, which allow you to isolate and find each variable's value.

It is crucial to understand that each equation represents a curve in a two-dimensional plane. The solutions will be the coordinate points where these curves intersect—essentially the points that lie on both curves simultaneously. By accurately applying methods like elimination, we can systematically reduce the complexity of these equations to find their intersection points.

Quadratic Equations

Quadratic equations are fundamental in algebra and are expressed in the general form \( ax^2 + bx + c = 0 \). In this context, we are dealing with quadratic equations embedded within a system, which adds some complexity. Quadratics, by nature, can have up to two solutions for \( x \) due to their parabolic shape when graphed.

The given system of equations has the following quadratic terms:

• \( 3x^2 \) in the first equation
• \( 2x^2 \) in the second equation

To find the solutions, we used the elimination method, which allowed us to isolate the \( x^2 \) term by eliminating the variable \( y \), eventually leading to a simplified quadratic equation. Solving \( 7x^2 = 77 \) by dividing by 7 gives \( x^2 = 11 \), and taking the square root results in the solutions \( x = \pm \sqrt{11} \).

During the problem-solving process, it’s important to correctly apply basic algebraic manipulations like factoring, expanding, and simplifying. Recognizing these operations and understanding their effect on quadratics is pivotal to mastering such problems.

Solving by Substitution

The substitution method is a powerful technique used to solve systems of equations. Although not directly used to solve this problem, understanding it can give more insight into how systems can be approached.

By substitution, one solves one of the equations for one variable in terms of the others and then substitutes this expression into the remaining equations. For instance, if we had used substitution and solved one equation for \( y \), we could then substitute this expression into the other equation. This results in simplifying the system into a single equation with one unknown, making it easier to solve.

In general, substitution is particularly useful when equations are amenable to getting one variable explicitly by itself without too much complexity. It was not the chosen method in our case, due to the quadratic terms. This choice highlights the elimination method’s suitability here, but being flexible and understanding multiple methods of solving systems always equips students to handle a variety of problems effectively.