Problem 12 - Precalculus Mathematics for Calculus

Question

Use the elimination method to find all solutions of the system of equations. $$\left\\{\begin{aligned} 2 x^{2}+4 y &=13 \\ x^{2}-y^{2} &=\frac{7}{2} \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solutions are \( \left( \pm\frac{5}{\sqrt{2}}, -3 \right) \) and \( \left( \pm\frac{3}{\sqrt{2}}, 1 \right) \).

Step 1: Choose a Variable to Eliminate

We need to choose one variable to eliminate from the given system of equations. Let's eliminate \(x^2\). The equations are:\[ 2x^2 + 4y = 13 \]\[ x^2 - y^2 = \frac{7}{2} \]

Step 2: Adjust the Equations

To eliminate \(x^2\), we can multiply the second equation by 2, so that both equations will have the same coefficient for \(x^2\). This gives:\[ 2(x^2 - y^2) = 2 \times \frac{7}{2} \]Simplifying, we get:\[ 2x^2 - 2y^2 = 7 \]

Step 3: Subtract the Equations

Subtract the modified second equation from the first equation to eliminate \(x^2\):\[ (2x^2 + 4y) - (2x^2 - 2y^2) = 13 - 7 \]Simplify the result:\[ 4y + 2y^2 = 6 \]

Step 4: Factor and Solve the Resulting Equation

Rearrange and simplify the equation:\[ 2y^2 + 4y - 6 = 0 \]Divide the entire equation by 2:\[ y^2 + 2y - 3 = 0 \]Factor the quadratic equation:\[ (y + 3)(y - 1) = 0 \]

Step 5: Solve for y

Set each factor equal to zero:\[ y + 3 = 0 \quad \rightarrow \quad y = -3 \] and\[ y - 1 = 0 \quad \rightarrow \quad y = 1 \]

Step 6: Solve for x Using Each y Value

Now substitute each value of \(y\) back into one of the original equations to solve for \(x\). Using the first original equation:For \(y = -3\):\[ 2x^2 + 4(-3) = 13 \]\[ 2x^2 - 12 = 13 \]\[ 2x^2 = 25 \]\[ x^2 = \frac{25}{2} \]\[ x = \pm \sqrt{\frac{25}{2}} \]\[ x = \pm \frac{5}{\sqrt{2}} \]For \(y = 1\):\[ 2x^2 + 4(1) = 13 \]\[ 2x^2 + 4 = 13 \]\[ 2x^2 = 9 \]\[ x^2 = \frac{9}{2} \]\[ x = \pm \sqrt{\frac{9}{2}} \]\[ x = \pm \frac{3}{\sqrt{2}} \]

Step 7: State All Solutions

Compile the solutions. For \(y = -3\), \(x = \pm\frac{5}{\sqrt{2}}\). For \(y = 1\), \(x = \pm\frac{3}{\sqrt{2}}\). Thus, the solutions are:- \( \left( \frac{5}{\sqrt{2}}, -3 \right) \)- \( \left( -\frac{5}{\sqrt{2}}, -3 \right) \)- \( \left( \frac{3}{\sqrt{2}}, 1 \right) \)- \( \left( -\frac{3}{\sqrt{2}}, 1 \right) \)

Key Concepts

System of Equations

A system of equations is a set of two or more equations that have common variables. Solving a system of equations means finding the values of these variables that satisfy all equations simultaneously. In our exercise, we have two equations:

• \(2x^2 + 4y = 13\)
• \(x^2 - y^2 = \frac{7}{2}\)

These are both quadratic equations, meaning they involve squared terms. When dealing with a system like this, the elimination method is a useful tool. This involves aligning the equations in such a way that adding or subtracting them will eliminate a variable. The aim is to simplify the system into equations with fewer variables, ideally leading you to find the variable values that satisfy all original equations. It's a step-by-step process that requires a bit of manipulation and is very handy for managing complex equations.

Quadratic Equations

Quadratic equations are polynomial equations of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) is an unknown. In simpler terms, these equations involve the square of a variable. Quadratics can look complicated, but they are quite manageable. In the context of our system of equations exercise, both equations contained terms with \(x^2\) and \(y^2\).
To easily work with them, you might need to factor or complete the square. In our solution steps, after elimination reduced one equation to a form where it could be rearranged as \(y^2 + 2y - 3 = 0\). This is a standard quadratic equation which we can solve by factoring into \((y + 3)(y - 1) = 0\). From here, solving for \(y\) becomes much simpler. Quadratics often have two solutions, hence finding distinct values for our variables.

Solution of Equations

Finding the solution to a system of equations means determining the values of the variables that make all equations true. Using methods like elimination, we simplify the equations step-by-step.
In our initial step, we chose to eliminate \(x^2\). By manipulating the equations to match coefficients and subtracting them, we reduced the problem to a simpler form. This allowed us to solve for \(y\), giving specific values. Once we know a value for one variable, substituting back into one of the original equations provides the value of the second variable.

• For \(y = -3\), we found \(x = \pm\frac{5}{\sqrt{2}}\)
• For \(y = 1\), we found \(x = \pm\frac{3}{\sqrt{2}}\)

Compiling these, the solution set gives us four pairs \((x, y)\). These are our solutions as they satisfy both equations in the system. Solving equations is about finding all possible values that make the initial conditions true.