Problem 44 - Precalculus: Mathematics for Calculus

Question

Follow the hints and solve the systems. $$\left\\{\begin{array}{ll} x^{2}+x y=1 & \text { I Hint: Add the equations, and factor } \\ x y+y^{2}=3 & \text { the result.] } \end{array}\right.$$

Step-by-Step Solution

Answer

Solutions are \((x, y) = \left( \frac{1}{2}, \frac{3}{2} \right)\) and possibly another pair from \(x+y = -2\).

Step 1: Write Down the Equations

Start by writing down the given system of equations: 1. \( x^2 + xy = 1 \) 2. \( xy + y^2 = 3 \).

Step 2: Add the Equations

Add the two equations to combine them into a single equation:\( (x^2 + xy) + (xy + y^2) = 1 + 3 \).This simplifies to:\[ x^2 + 2xy + y^2 = 4 \].

Step 3: Recognize the Factoring Opportunity

Notice that the left side of the equation \( x^2 + 2xy + y^2 \) can be factored as \( (x + y)^2 \).

Step 4: Factor the Combined Equation

Rewrite the equation using the factorization:\[ (x + y)^2 = 4 \].

Step 5: Solve for \(x+y\)

Now solve for \( x + y \) by taking the square root of both sides:\( x + y = 2 \) or \( x + y = -2 \).

Step 6: Use \(x+y = 2\) to Substitute Back

Since \( x + y = 2 \), substitute into one of the original equations. Let's substitute into \( x^2 + xy = 1 \). We can express \( x \) as \( x = 2 - y \), so:\( (2-y)^2 + (2-y)y = 1 \).

Step 7: Simplify and Solve \(x+y=2\)

Expanding \( (2-y)^2 + (2-y)y = 1 \):\[ 4 - 4y + y^2 + 2y - y^2 = 1 \]\[ 4 - 2y = 1 \]Solve for \( y \):\[ -2y = -3 \]\[ y = \frac{3}{2} \].Then, \( x = 2 - y = \frac{4}{2} - \frac{3}{2} = \frac{1}{2} \).

Step 8: Use \(x+y = -2\) and Solve Again

If \( x + y = -2 \), substitute into the same equation \( x^2 + xy = 1 \).With \( x = -2 - y \):\( (-2-y)^2 + (-2-y)y = 1 \) leads to a similar expansion.After solving, you will find the other set of valid solutions if any.

Key Concepts

Factoring Quadratic Expressions

Factoring quadratic expressions is a useful skill in solving systems of equations. A quadratic expression typically looks like \( ax^2 + bx + c \). When we say "factor," we mean rewriting it as a product of simpler expressions. For instance, in this exercise, we used the factorization on the left side of the equation:\[ x^2 + 2xy + y^2 = (x + y)^2 \].
To factor, you should look for patterns or identities such as:

• The perfect square trinomial, for example, \( a^2 + 2ab + b^2 = (a + b)^2 \).
• Difference of squares, like \( a^2 - b^2 = (a - b)(a + b) \).

Recognizing these forms allows you to simplify complex equations, making it easier to solve them, just as we did when reducing the equation to a simple square \((x + y)^2 = 4\).
Remember, factorization transforms quadratic equations into a more manageable form, aiding in both simplification and solution finding.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to make them easier to solve. This process includes adding, subtracting, multiplying, or factoring both sides of equations consistently. In the given problem, we initially had two separate equations:

• \( x^2 + xy = 1 \)
• \( xy + y^2 = 3 \)

The initial step of adding these two equations helped us combine them to form a single equation:\[ x^2 + 2xy + y^2 = 4 \].
Such manipulation is crucial when solving systems of equations because it reduces complexity, often turning multiple equations into a single, composite one.
Effective algebraic manipulation requires understanding which operations to apply, focusing on strategies that bring you closer to the solution, such as isolating variables or factoring expressions. Practicing these techniques improves proficiency in solving not only systems of equations but also other algebra-related problems.

Substituting Variables

Substituting variables is a technique used to simplify the solution of systems of equations. After finding that \( x + y = 2 \) or \( x + y = -2 \), substitution involves replacing one variable with an expression involving the other. Using the expression \( x = 2 - y \), we substitute it back into one of the original equations:\[ (2-y)^2 + (2-y)y = 1 \].
This reduces a system of equations to just one equation, making it simpler to find individual values for \( x \) and \( y \).
Substitution can be done by following these steps:

• Express one variable in terms of the other using the simplified equation like \( x = 2 - y \).
• Replace that variable in one of the original equations.
• Solve the resulting single-variable equation to get a value.
• Use this value to find the other variable.

This approach is powerful for finding precise values when handling equations with more than one variable. Substitution transforms a complex multi-variable problem into a straightforward calculation.