Question

Solve each system. Use any method you wish. $$ \left\\{\begin{array}{r} 7 x^{2}-3 y^{2}+5=0 \\ 3 x^{2}+5 y^{2}=12 \end{array}\right. $$

Short answer

The solutions are \( (\frac{1}{2}, \frac{3}{2}), (-\frac{1}{2}, \frac{3}{2}), (\frac{1}{2}, -\frac{3}{2}), (-\frac{1}{2}, -\frac{3}{2}) \).

Step-by-step solution

Identify the System of Equations

The system of equations to be solved is: \[ \begin{cases} 7x^{2} - 3y^{2} + 5 = 0 \ 3x^{2} + 5y^{2} = 12 \ \text{(1)} \ \text{(2)} \ \text{We will solve this system using the Substitution Method.} \end{{cases}} \]

Solve for One Variable in Terms of the Other

First, rearrange Equation (1) to solve for one variable in terms of the other. \[ \ 7x^{2} = 3y^{2} - 5 \ x^{2} = \frac{3y^{2} - 5}{7} \ \text{(3)} \]

Substitute into the Second Equation

Substitute the expression for \(x^{2}\) from Equation (3) into Equation (2). \[ \ 3 \times \frac{3y^{2} - 5}{7} + 5y^{2} = 12 \ \frac{9y^{2} - 15}{7} + 5y^{2} = 12 \ 9y^{2} - 15 + 35y^{2} = 84 \ 44y^{2} - 15 = 84 \ 44y^{2} = 99 \ y^{2} = \frac{99}{44} = \frac{9}{4} = 2.25 \ y^{2} = 2.25 \ y = \frac{3}{2} \text{ or } y = -\frac{3}{2} \]

Solve for the Corresponding Values of x

Using the values of \(y\), we solve for \(x\) using Equation (3). For \( y = \frac{3}{2} \): \[ x^{2} = \frac{3(\frac{9}{4}) - 5}{7} = \frac{\frac{27}{4} - 5}{7} = \frac{\frac{27}{4} - \frac{20}{4}}{7} = \frac{7/4}{7} = \frac{1}{4} \ x = \frac{1}{2} \text{ or } x = -\frac{1}{2} \] For \( y = -\frac{3}{2} \): \[ x^{2} = \frac{3(\frac{9}{4}) - 5}{7} = \frac{\frac{27}{4} - 5}{7} = \frac{\frac{27}{4} - \frac{20}{4}}{7} = \frac{7/4}{7} = \frac{1}{4} \ x = \frac{1}{2} \text{ or } x = -\frac{1}{2} \]

Write the Solution Set

Combine the values of \(x\) and \(y\) from the previous steps to write the solution set. \[ \{ (\frac{1}{2}, \frac{3}{2}), (-\frac{1}{2}, \frac{3}{2}), (\frac{1}{2}, -\frac{3}{2}), (-\frac{1}{2}, -\frac{3}{2}) \} \]

Key concepts

Substitution Method

The substitution method is a common technique for solving systems of equations. It involves solving one of the equations for one variable, and then substituting that expression into the other equation. This method is especially useful when dealing with nonlinear equations as it simplifies the system into a single equation with one variable. In our exercise, we began by rearranging the first equation to express \(x^2\) in terms of \(y^2\). Then, we substituted this expression into the second equation. This helped reduce the problem to a simpler form, allowing us to solve for the variables step-by-step. Ensuring accuracy at this stage is crucial since any mistake can affect the final solution set.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to isolate variables. In the given exercise, algebraic manipulation was used extensively to rearrange and simplify the equations. After expressing \(x^2\) from the first equation, we needed to substitute it into the second equation. This required careful handling of fractions and combining like terms. Another key aspect was balancing the equations correctly through addition and multiplication. Simplifying fractions and understanding the properties of equality helps in maintaining accuracy throughout the steps.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). In our problem, both equations were quadratic in form with terms involving \(x^2\) and \(y^2\). Solving quadratic equations often involves finding the square roots of variables, as seen when we isolated \(y^2\). Then, knowing that \(y^2 = 2.25\), we found that \(y\) could be \(\frac{3}{2}\) or \(-\frac{3}{2}\). It’s important to consider both positive and negative roots when solving quadratic equations, as each provides potential solutions for the system. Remember to verify all potential solutions in the original equations to ensure they are valid.

Solution Set

The solution set of a system of equations includes all ordered pairs \((x, y)\) that satisfy both equations. In this case, the solution set was found by solving for \(y\) and then finding the corresponding values of \(x\). Our final solution set contained four pairs: \((\frac{1}{2}, \frac{3}{2})\), \((-\frac{1}{2}, \frac{3}{2})\), \((\frac{1}{2}, -\frac{3}{2})\), and \((-\frac{1}{2}, -\frac{3}{2})\). These pairs were derived by considering both positive and negative root values of \(x\) and \(y\). It's crucial to present the solution set in an organized way, ensuring that all unique solutions are included. This helps in visualizing the intersection points of graphs represented by the equations, confirming the completeness of the solution.