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

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

Step 1: Identify the elimination strategy

The given system of equations is \(3x^2 + 4y = 17\) and \(2x^2 + 5y = 2\). To use elimination, we should eliminate one variable. It's efficient to eliminate \(y\) by equalizing the coefficients of \(y\) in both equations.

Step 2: Adjust equations for elimination

To eliminate \(y\), multiply the first equation by \(5\) (the coefficient of \(y\) in the second equation), and the second equation by \(4\) (the coefficient of \(y\) in the first equation) to obtain:1. \(15x^2 + 20y = 85\)2. \(8x^2 + 20y = 8\).

Step 3: Eliminate the variable

Subtract the second adjusted 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 \(x\)

Divide both sides of \(7x^2 = 77\) by \(7\) to find:\(x^2 = 11\).Take the square root of both sides, yielding \(x = \sqrt{11}\) or \(x = -\sqrt{11}\).

Step 5: Substitute \(x\) to find \(y\)

Using \(x = \sqrt{11}\), substitute back into the original equation, suppose the first:\(3(\sqrt{11})^2 + 4y = 17\), which simplifies to:\(33 + 4y = 17\).Solve for \(y\):\(4y = 17 - 33\), so \(4y = -16\) and \(y = -4\).Now use \(x = -\sqrt{11}\), substitute similarly:\(3(-\sqrt{11})^2 + 4y = 17\), again \(33 + 4y = 17\) resulting in:\(y = -4\).

Step 6: Solution confirmation

The solutions are verified by substituting \((x,y) = (\sqrt{11}, -4)\) and \((x,y) = (-\sqrt{11}, -4)\) into both original equations. They both satisfy the systems, confirming the correctness of the solutions.

Key Concepts

System of Equations

A system of equations consists of two or more equations with the same set of variables. The main objective is to find values for these variables that satisfy all the given equations simultaneously. In a system of equations, each equation acts as a constraint on the possible values of the variables.

For example, in the given system of equations:

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

we are asked to find values of \(x\) and \(y\) that make both equations true at the same time. Solving systems of equations can be done through various methods, including substitution, graphing, and elimination, each with its strengths based on the problem at hand.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. The equation can have up to two real solutions. In our system, the variables \(x^2\) suggest that these are quadratic equations because of the squared term.

In the process of solving the system, even though there is no explicit quadratic equation on its own to solve directly, the presence of \(x^2\) necessitates understanding of solving quadratics when isolating variables.

The method involves manipulating the system until one of the equations is reduced in such a way that allows one to set up a typical quadratic form, often using substitution or elimination.

Variable Elimination

Variable elimination is a straightforward technique used to simplify solving systems of equations by removing one variable, making it easier to solve for the other. This is done by aligning the equations so that one variable cancels out, leaving an equation in a single variable.

In our problem, we choose to eliminate the variable \(y\) by manipulating the equations. We multiply each equation to make the coefficients of \(y\) equal so that when the equations are subtracted, \(y\) disappears.

• For the first equation, multiply by \(5\) to get: \(15x^2 + 20y = 85\)
• For the second equation, multiply by \(4\) to get: \(8x^2 + 20y = 8\)

By subtracting the second equation from the first, \(y\) is eliminated, and you are left with a simpler equation \(7x^2 = 77\), facilitating an easy solution for \(x\). Once \(x\) is determined, substitute back to find \(y\). This efficient method drastically reduces the complexity of solving the system.