Problem 56 - College Algebra with Modeling and Visualization

Question

If possible, solve the nonlinear system of equations. $$ \begin{aligned} &x^{2}-y=3\\\ &x+y=3 \end{aligned} $$

Step-by-Step Solution

Answer

The solutions are \((-3, 6)\) and \((2, 1)\).

Step 1: Express y in terms of x

Start by solving the linear equation for one of the variables. From the equation \(x + y = 3\), express \(y\) in terms of \(x\):\[ y = 3 - x \]

Step 2: Substitute y in the nonlinear equation

Substitute the expression for \(y\) from Step 1 into the nonlinear equation \(x^2 - y = 3\):\[ x^2 - (3 - x) = 3 \]

Step 3: Simplify the equation

Simplify the equation obtained from Step 2:\[x^2 - 3 + x = 3\]Subtract 3 from both sides:\[x^2 + x - 6 = 0\]

Step 4: Factor the quadratic equation

Factor the quadratic equation \(x^2 + x - 6 = 0\):\[(x + 3)(x - 2) = 0\]

Step 5: Solve for x

Solve the equation \((x + 3)(x - 2) = 0\) for \(x\):\[x + 3 = 0 \quad \text{or} \quad x - 2 = 0\]This leads to the solutions:\[x = -3 \quad \text{or} \quad x = 2\]

Step 6: Solve for y

Use the expressions for \(x\) from Step 5 to find the corresponding \(y\) values using \(y = 3 - x\):If \(x = -3\), then \(y = 3 - (-3) = 6\).If \(x = 2\), then \(y = 3 - 2 = 1\).

Step 7: Write the solutions

The solutions for the system of equations are the pairs:\[(-3, 6) \quad \text{and} \quad (2, 1)\]

Key Concepts

Solving Quadratic Equations

At the heart of many algebra problems lies the quadratic equation. Quadratic equations come in the standard form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\). There are different ways to solve them, including factoring, using the quadratic formula, and completing the square.When dealing with quadratic equations, it's crucial to be clear about:

• Identifying the right form: Rearrange everything to one side so it equals zero.
• Understanding the roots: These are the values of \(x\) that satisfy the equation.
• Choosing a method: Depending on the specific problem, some methods could be more efficient than others.

In the problem from the exercise, solving the quadratic \(x^2 + x - 6 = 0\) is pivotal to finding the solutions of the system. A significant observation is that not all quadratics factor neatly; sometimes, applying the quadratic formula is the best recourse. However, when they do factor, it simplifies the process greatly, as it did in this instance.

Factoring Quadratics

Factoring is a technique where we express a quadratic equation as a product of simpler expressions. This process hinges on ensuring the quadratic can be rewritten in a simple form like \((x + p)(x + q) = 0\), where \(p\) and \(q\) are numbers that add up to the coefficient of \(x\) and multiply to the constant term.Here's what factoring involves:

• Recognize a pattern: Especially with ax² + bx + c, look for two numbers that multiply to ac (the product of \(a\) and \(c\)) and add to \(b\).
• Rewriting: Break the middle term using these numbers and factor by grouping if necessary.
• Solve: Set each factor equal to zero to find the solutions.

In the exercise solution, the equation \(x^2 + x - 6 = 0\) was factored into \((x + 3)(x - 2) = 0\), a step that directly offered solutions for \(x\). Factoring simplifies the equation and reveals the values that satisfy it, making it a powerful tool in your algebra toolbox.

Algebraic Substitution

Algebraic substitution is a technique used to replace a variable with an equivalent expression. It is especially useful in solving systems of equations where one equation can be manipulated to express one variable in terms of another. The resulting expression is then substituted into the other equation(s).Here's how it works:

• Solve for a variable: Use the simplest equation to express one variable in terms of the other.
• Substitute: Replace this variable in the other equation with the expression found.
• Simplify and Solve: The new equation will now have one variable, making it solvable.

In the provided system, the linear equation \(x + y = 3\) was manipulated to express \(y\) as \(3 - x\). This expression was then substituted in the nonlinear equation \(x^2 - y = 3\), resulting in a single-variable equation that could then be simplified and solved. Such substitution simplifies complex systems, making solutions more apparent.