Problem 18 - College Algebra Essentials

Question

In Exercises \(1-18,\) solve each system by the substitution method. $$ \left\\{\begin{array}{l} 2 x+y=4 \\ (x+1)^{2}+(y-2)^{2}=4 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is {(1, 2), (0.2, 3.6)}.

Step 1: Isolate a variable from the linear equation

From the equation \(2x + y = 4\), isolate \(y\) to get \(y = 4 - 2x\).

Step 2: Substitute in quadratic equation

Replace \(y\) with \(4 - 2x\) in the equation \((x+1)^2 + (y - 2)^2 = 4\), which becomes \((x + 1)^2 + ((4 - 2x) - 2)^2 = 4\). Simplify this to get the quadratic equation \(x^2 - 2x + 1 + 4x^2 -16x + 4 = 4\). Collecting like terms, we get \(5x^2 - 18x + 1 = 0\).

Step 3: Solve the quadratic equation

Solve \(5x^2 - 18x + 1 = 0\) using the quadratic formula. The solutions for \(x\) are \(x=1\) and \(x=0.2\).

Step 4: Find corresponding y values

Substitute \(x=1\) and \(x=0.2\) separately in the equation \(y = 4 - 2x\), we get \(y = 4 - 2*1 = 2\) for \(x=1\) and \(y = 4 - 2*0.2 = 3.6\) for \(x=0.2\).

Key Concepts

System of Equations

When dealing with a system of equations, you are essentially looking at two or more equations that are set equal to each other with the goal of finding the value(s) of the variables that satisfy all of the equations simultaneously. There are multiple methods to tackle these systems, including graphing, elimination, and substitution.

The substitution method is highly useful when one of the equations can be manipulated to isolate one variable. This isolated variable can then be substituted into the other equation(s), reducing the system to a single equation with one variable that can be easily solved. It's a step-by-step process that requires careful manipulation and often, simplification of the equations involved.

For instance, in the given exercise, we first isolated one variable from the linear equation and then substituted it into a quadratic equation. This method not only provides a systematic approach to finding solutions but also helps in developing a deeper understanding of the interplay between linear and quadratic functions within a system.

Quadratic Equations

A quadratic equation is a second-degree polynomial equation typically in the form of \( ax^2 + bx + c = 0 \), where \( a \) is not equal to zero. The solutions to these equations can be found using methods such as factoring, completing the square, or applying the quadratic formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Quadratics are recognizable due to their curved graphs, known as parabolas, and they can have zero, one, or two real solutions depending on the value of the discriminant \( (b^2 - 4ac) \). In the exercise, the substitution method transformed the system into a quadratic equation, which was then solved using the quadratic formula. This highlights the usefulness of having diverse tools to analyze and solve equations of different complexities.

Solving Linear Equations

Solving linear equations is foundational for algebra and represents an immense field of practical application, ranging from simple budgeting to complex engineering problems. A linear equation is an equation of the first degree, meaning it contains no exponents higher than one and graphs as a straight line.

The general form of a linear equation in two variables is \( ax + by = c \), where \( a \) and \( b \) are coefficients, \( x \) and \( y \) are the variables, and \( c \) is the constant. Solving a linear equation entails finding the value of the variable that will make the equation true.

In our example, solving the linear equation involved isolating \( y \) to express it in terms of \( x \) and then substituting this expression into another equation. This is a crucial skill in algebra, as it allows for the simplification and resolution of more complicated systems.