Problem 39 - College Algebra

Question

Solve each system by the method of your choice. $$\begin{aligned} &y=(x+3)^{2}\\\ &x+2 y=-2 \end{aligned}$$

Step-by-Step Solution

Answer

The solutions to the system of equations are \((-2, 1)\) and \((-5, 4)\)

Step 1: Identify the substitution variable

Since the first equation \(y = (x+3)^{2}\) expresses y as a function of x, we can substitute \((x+3)^{2}\) into y in the second equation.

Step 2: Substitute y in the second equation

Substitute \((x+3)^{2}\) in place of y in the second equation, resulting to: \(x+2(x+3)^{2}=-2\)

Step 3: Simplify the resulting equation and solve for x

We get \(x + 2(x^2 + 6x + 9) = -2\), which is equivalent to \(x + 2x^2 + 12x + 18 = -2\). Simplify this further to: \(2x^2 + 13x + 20 = 0\). Now, solve this quadratic equation for x. The solutions are \(x = -2, -5\)

Step 4: Substitute x into first equation and solve for y

Substitute \(x = -2\) and \(x = -5\) back in the first equation \(y = (x+3)^{2}\) to get the corresponding y values. When \(x = -2\), we have \(y = (-2+3)^{2} = 1\) and when \(x = -5\), we have \(y = (-5+3)^{2} = 4\)

Step 5: Summary and presentation of solutions

So, the solutions to the system of equations are \((-2, 1)\) and \((-5, 4)\)

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of equations. It involves an initial step of expressing one variable in terms of the other. In our exercise, the equation given is \(y = (x+3)^2\). This equation already shows \(y\) in terms of \(x\), making it perfect for substitution. By substituting the expression for \(y\) into the other equation, we create an equation with just one variable. This simplifies the problem and helps to solve it piece by piece, bridging from a system of equations to a single equation. This method is ideal for systems where one equation is particularly simple to rearrange to express one variable. It provides a straightforward path to finding solutions through systematic elimination of variables.

Quadratic Equation

A quadratic equation is any equation of the form \(ax^2 + bx + c = 0\). In our problem, after using the substitution method, we simplified the system into the quadratic equation \(2x^2 + 13x + 20 = 0\). Such equations are critically important in mathematics as they emerge in various real-world phenomena. Solving these equations typically involves finding \(x\) values that satisfy this equation. These solutions can be found using different methods, including factoring, completing the square, or using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). In this exercise, factoring brings us to the solutions \(x = -2\) and \(x = -5\), each representing possible values that satisfy the original quadratic equation.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to make them more manageable. In the exercise, after substituting, the equation \(x + 2(x+3)^2 = -2\) needs restructuring. Expanding \((x+3)^2\) to \(x^2 + 6x + 9\) turns the equation into \(x + 2x^2 + 12x + 18 = -2\). The next step is collecting like terms, leading to the simplified form \(2x^2 + 13x + 20 = 0\). This kind of manipulation requires careful balancing of the equation while performing operations on both sides, keeping everything equal. It ensures each step progresses towards a simpler form that is easier to work with until reaching a solvable equation.

Polynomial Equations

A polynomial equation consists of variables raised to whole number exponents, which are summed together. In our example, the derived equation \(2x^2 + 13x + 20 = 0\) is a polynomial of degree 2, known as a quadratic polynomial. These equations can have as many degrees as the term with the highest exponent indicates, affecting the number and type of solutions. Handling polynomial equations involves techniques like factoring, division, or applying the quadratic formula for solving quadratic polynomials specifically. Understanding polynomial equations is crucial because they form the foundation for higher mathematics and are deeply rooted in describing natural phenomena and solving practical problems.