Problem 63 - Intermediate Algebra

Question

Solve each system of equations for real values of \(x\) and \(y.\) $$ \left\\{\begin{array}{l} x^{2}-y^{2}=4 \\ x+y=4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{5}{2}\) and \(y = \frac{3}{2}\).

Step 1: Solve for One Variable

Start with the second equation: \(x + y = 4\). Solve for one variable, let's solve for \(y\):\[ y = 4 - x. \]

Step 2: Substitute and Simplify

Substitute \(y = 4 - x\) into the first equation \(x^2 - y^2 = 4\):\[ x^2 - (4 - x)^2 = 4. \] Simplify using the identity \((a-b)^2 = a^2 - 2ab + b^2\):\[ x^2 - (16 - 8x + x^2) = 4. \] Simplify further to eliminate \(x^2\):\[ -16 + 8x = 4. \]

Step 3: Solve for Variable

Solve the equation \( -16 + 8x = 4\) for \(x\):\[ 8x = 20. \] Divide both sides by 8 to find \(x\):\[ x = \frac{20}{8} = \frac{5}{2}. \]

Step 4: Substitute Back to Find y

Substitute \(x = \frac{5}{2}\) back into \(y = 4 - x\):\[ y = 4 - \frac{5}{2} = \frac{8}{2} - \frac{5}{2} = \frac{3}{2}. \]

Step 5: Verify the Solution

Verify the solution by checking both original equations. For \(x^2 - y^2 = 4\): \[(\frac{5}{2})^2 - (\frac{3}{2})^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4.\] For \(x + y = 4\): \[\frac{5}{2} + \frac{3}{2} = \frac{8}{2} = 4.\] Both equations are satisfied, confirming the solution is correct.

Key Concepts

Quadratic Equations

Quadratic equations are one of the fundamental types of polynomial equations. They are called quadratic because the highest power of the variable is two. In the context of this exercise, one such quadratic equation is given as follows:

• \(x^2 - y^2 = 4\)

Understanding how to work with them is crucial as they appear frequently in various forms of mathematics and applications. Whenever you encounter a quadratic equation, remember these key features:

• The graph of a quadratic equation is a parabola, which can open upwards or downwards.
• The solutions to a quadratic equation are called roots. You may find them using methods like factoring, completing the square, or the quadratic formula.

In this exercise, the quadratic part comes into play when we implement a method called the substitution method, which helps us solve the simultaneous equations. Solving quadratic equations often requires algebraic manipulation to simplify them and find the variable values.

Algebraic Manipulation

Algebraic manipulation is the process of rearranging and simplifying equations to make it easier to solve for unknowns. It is a critical skill when dealing with systems of simultaneous equations like those in the problem. Here’s why algebraic manipulation is so important:

• It helps in isolating terms, which is crucial for extracting values of unknown variables.
• Applying algebraic identities, such as \((a-b)^2 = a^2 - 2ab + b^2\), simplifies complex expressions into manageable terms.
• Operations like addition, subtraction, multiplication, division, and factoring are used frequently.

In the solution steps, after substituting \(y = 4 - x\) into the quadratic equation, algebraic manipulation was used to eliminate the \(x^2\) terms, simplifying it to \(-16 + 8x = 4\). Then, by rearranging, we found a clear path to finding \(x\). This step-by-step manipulation helps untangle the equations so we can solve them efficiently.

Substitution Method

The substitution method is a technique used to solve a system of equations. The idea is to solve one of the equations for one variable and then substitute that expression into the other equation. This method is very effective when dealing with linear and quadratic systems as in this problem. Let's explore why the substitution method is useful:

• It reduces the number of variables in an equation, simplifying the solving process.
• It allows us to convert a system into a single equation, often making it easier to manage.
• It is particularly useful when one equation in the system is a simple expression, such as \(x+y=4\) in this exercise.

In this particular problem, the substitution method was initiated by taking the simpler linear equation \(x + y = 4\) and solving it for \(y\) as \(y = 4 - x\). By inserting this expression for \(y\) into the quadratic equation, we converted a two-variable problem into a single-variable problem. This step is often the key to simplifying and solving complex systems in algebra.