Problem 46 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}+y^{2}=10 \\ 2 x^{2}-3 y^{2}=5 \end{array}\right. $$

Step-by-Step Solution

Answer

The solutions for \((x, y)\) are: \((\sqrt{7}, \sqrt{3})\), \((-\sqrt{7}, \sqrt{3})\), \((\sqrt{7}, -\sqrt{3})\), \((-\sqrt{7}, -\sqrt{3})\)."

Step 1: Identify and Set Up the Equations

We begin with the given system of equations: 1. \( x^2 + y^2 = 10 \) 2. \( 2x^2 - 3y^2 = 5 \) We need to solve for the real values of \(x\) and \(y\) that satisfy both equations simultaneously.

Step 2: Solve for One Variable in One Equation

From Equation (1): \[ x^2 + y^2 = 10 \] We can express one variable in terms of the other. Let's solve for \(x^2\): \[ x^2 = 10 - y^2 \] We'll substitute this expression into Equation (2).

Step 3: Substitute and Simplify

Substitute \(x^2 = 10 - y^2\) from Step 2 into Equation (2):\[ 2(10 - y^2) - 3y^2 = 5 \] This simplifies to: \[ 20 - 2y^2 - 3y^2 = 5 \] Further combining like terms: \[ 20 - 5y^2 = 5 \]

Step 4: Solve for \(y\)

Rearrange the equation: \[ 20 - 5 = 5y^2 \]\[ 15 = 5y^2 \] Divide both sides by 5: \[ y^2 = 3 \] So, \( y = \pm \sqrt{3} \).

Step 5: Solve for \(x\) Using \(y\) Values

Now we have two possible \(y\) values: \( y = \sqrt{3} \) and \( y = -\sqrt{3} \). Substitute these back into \( x^2 = 10 - y^2 \):1. If \( y = \sqrt{3} \): \[ x^2 = 10 - (^3) = 7 \] \( x = \pm \sqrt{7} \)2. If \( y = -\sqrt{3} \): \[ x^2 = 10 - (^3) = 7 \] \( x = \pm \sqrt{7} \).

Step 6: Write Down All Possible Solutions

From Step 5, we have the possible solution pairs:1. \((x, y) = (\sqrt{7}, \sqrt{3})\)2. \((x, y) = (-\sqrt{7}, \sqrt{3})\)3. \((x, y) = (\sqrt{7}, -\sqrt{3})\)4. \((x, y) = (-\sqrt{7}, -\sqrt{3})\).

Key Concepts

Real Values

When solving equations, particularly those in a system like the one given, looking for "real values" means we are interested in finding solutions that are real numbers. These are numbers that include all the numbers on the number line and they can be positive, negative, or zero. Real numbers do not include imaginary numbers, which involve the square root of negative numbers (like \(\sqrt{-1}\), often represented by \(i\)).

In this exercise, the goal is to find the real values of \(x\) and \(y\) that satisfy both equations in the system simultaneously. Solutions like \(x = \sqrt{7}\) or \(y = \sqrt{3}\) are considered real values because they are positive numbers we can locate on the number line.

It's important to make sure that the solutions are sensible within the context of real-world numbers. If during solving you end up with a square root of a negative number, then it does not result in a real value, and you'd need to reassess how to approach the system of equations.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(x\) represents an unknown variable. Quadratic equations can have up to two real solutions. These occur when the polynomial is equal to zero.

In the given exercise, both equations in the system are quadratic because they involve \(x^2\) and \(y^2\) terms. Quadratic equations can be solved using several methods:

• Factoring: Rewriting the quadratic as a product of two binomials.
• Quadratic formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find solutions.
• Completing the square: Rearranging and simplifying the quadratic expression to make finding roots easier.

In the exercise, a substitution method was used where one quadratic equation was solved for one variable and substituted into the other. This approach systematically reduces complexity and allows us to solve for one variable at a time.

Simultaneous Equations

Simultaneous equations involve finding values for variables that satisfy all given equations at the same time. In this context, they provide a set of conditions each variable must meet.

There are several techniques for solving simultaneous equations, especially when dealing with linear or quadratic forms:

• Substitution: Solve one equation for a variable and replace it in the other equation, as was done in this example.
• Elimination: Add or subtract equations to eliminate a variable, simplifying the process of finding solutions.
• Graphical Method: Plotting the equations on a graph to find intersections, representing solutions visually.

In this particular problem, by expressing \(x^2\) in terms of \(y^2\) using the substitution method, it allowed for simplifying the system and solving for one variable effectively. Once one variable was known, this insight allowed us to determine the value of the other variable, ensuring that the pairs of solutions we end up with satisfy the original conditions set by both equations.