Problem 28 - College Algebra

Question

For the following exercises, use any method to solve the nonlinear system. $$-x^{2}+y=2$$ $$-4 x+y=-1$$

Step-by-Step Solution

Answer

Solutions are (3, 11) and (1, 3).

Step 1: Solve the second equation for y

The second equation is \(-4x + y = -1\). We can express \(y\) in terms of \(x\) by adding \(4x\) to both sides of the equation, yielding:\[y = 4x - 1\]

Step 2: Substitute y in the first equation

Now that we have \(y\) in terms of \(x\), substitute \(y = 4x - 1\) into the first equation, which is \(-x^2 + y = 2\).This results in:\[-x^{2} + (4x - 1) = 2\] which simplifies to: \[-x^2 + 4x - 1 = 2\].

Step 3: Simplify and solve for x

From the equation \(-x^2 + 4x - 1 = 2\), add \(1\) to \(2\): \[-x^2 + 4x - 3 = 0\]. Multiply the entire equation by \(-1\) to simplify:\[x^2 - 4x + 3 = 0\]. Now, solve the quadratic equation by factoring:\[(x - 3)(x - 1) = 0\]. This gives the solutions: \(x = 3\) or \(x = 1\).

Step 4: Find corresponding y values

Substitute \(x = 3\) into the equation \(y = 4x - 1\):\[y = 4(3) - 1 = 11\].So one solution is \((3, 11)\).Substitute \(x = 1\) into \(y = 4x - 1\):\[y = 4(1) - 1 = 3\].So another solution is \((1, 3)\).

Step 5: Verify the solutions

Check each pair by substituting back into the original equations:1. For \((3, 11)\), Substitute into \(-x^2 + y = 2\):\(-3^2 + 11 = -9 + 11 = 2\) ✔️Substitute into \(-4x + y = -1\):\(-4(3) + 11 = -12 + 11 = -1\) ✔️2. For \((1, 3)\),Substitute into \(-x^2 + y = 2\):\(-1^2 + 3 = -1 + 3 = 2\) ✔️Substitute into \(-4x + y = -1\):\(-4(1) + 3 = -4 + 3 = -1\) ✔️Both pairs satisfy the original system.

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of degree two. The standard form of a quadratic equation is written as: \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(a eq 0\). These equations can be recognized by the squared variable, indicating they have a maximum of two solutions, known as roots.
Solving quadratic equations is a fundamental skill in algebra. The most common methods include:

• Factoring: Expressing the quadratic in a product of binomials.
• Quadratic Formula: Using the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).
• Completing the Square: Rewriting the quadratic as a perfect square trinomial.

The solutions to a quadratic equation visually represent the points where a parabola intersects the x-axis.
In our problem, after substituting values, we simplified to the quadratic equation \(x^2 - 4x + 3 = 0\), which we solved by factoring. Understanding how to manipulate and solve quadratics is vital as it's frequently used in various math topics.

Substitution Method

The substitution method is a powerful technique used for solving systems of equations. This method involves solving one equation for one variable and substituting this expression into the other equation.
This helps to transform two equations into a single equation, which is often easier to solve. The steps are typically as follows:

• Solve one equation for one variable in terms of the other variable(s).
• Substitute this expression into the other equation(s).
• Solve the new equation for the remaining variable.
• Use the obtained solution to find the value of the other variable.

In the original exercise, we used the substitution method by expressing \(y\) in terms of \(x\) from the second equation and substituting it into the first. This led to simplifying and eventually solving a single quadratic equation.
Substitution is especially helpful in systems with linear and nonlinear equations or when one variable is easily isolated.

Factoring

Factoring is a process used to simplify expressions or solve equations by breaking a composite number or polynomial down into products of its factors. It's akin to finding the ingredients that multiply together to yield the original expression.
When applied in solving quadratic equations, like \(ax^2 + bx + c = 0\), the goal is to express it as \((px + q)(rx + s) = 0\). This form allows finding the values of \(x\) for which the equation holds, because if a product of factors equals zero, at least one of the factors must be zero.
In the problem solution, we factored \(x^2 - 4x + 3\) as \((x - 3)(x - 1)\). Once factored, we quickly identified that \(x = 3\) or \(x = 1\) satisfy the equation, providing solutions to the system. Factoring not only simplifies solving quadratic equations but also aids in understanding the relationships between different elements of the polynomial.

Systems of Equations

Systems of equations consist of two or more equations with the same variables. Solving them means finding values for the variables that satisfy all equations simultaneously.
There are various methods to solve systems of equations, including:

• Graphical Method: Plotting each equation on a graph and finding their intersection points.
• Substitution Method: As previously explained, involves substituting one equation into another.
• Elimination Method: Adding or subtracting equations to eliminate a variable.

In the exercise, we faced a nonlinear system, as the presence of \(-x^2 + y = 2\) included a quadratic term. These systems can be more challenging than linear ones because they may have multiple solutions or require specific techniques like substitution or graphical interpretation to solve.
Understanding how each equation interacts within a system is crucial for finding the correct solution. In our solved exercise, we effectively used substitution to transform it into a solvable quadratic form, demonstrating the importance of choosing appropriate methods depending on the system's nature.