Problem 20 - College Algebra Essentials

Question

In Exercises \(19-30,\) solve each system by the addition method. $$ \left\\{\begin{array}{l} x+y=1 \\ x-y=3 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the given system of equations is \(x=2\) and \(y=-1\).

Step 1: Addition of the two equations

Begin by adding both equations together. This is done by adding the left sides and the right sides separately. Thus, \( (x+y) + (x-y) = 1 + 3\).\nThis simplifies to \( 2x = 4 \) since the \(y\) terms cancel each other out.

Step 2: Solve for 'x'

With the simplified equation \(2x = 4\), divide both sides by 2 to find the value of \(x\). This gives \(x = 4/2 = 2\)

Step 3: Find the value for 'y'

Substitute \(x = 2\) into one of the original equations to solve for \(y\). Substituting \(x\) into the first equation gives: \(2 + y = 1\). Then, subtract 2 from both sides of the equation to isolate \(y\). Thus, the value of \(y\) is \(1 - 2 = -1\).

Key Concepts

Addition Method

The addition method is a popular technique for solving systems of linear equations. It involves adding two equations together to eliminate one of the variables, making it easier to solve the system step by step. The goal is to align both equations in such a way that one variable cancels out when the equations are added. This helps to simplify the problem significantly.

How does it work? When you have a system like:

• Equation 1: \( x + y = 1 \)
• Equation 2: \( x - y = 3 \)

you align the terms with similar variables together and add up both sides of the equations. In our case, the \( y \) terms cancel out because they are the same but with opposite signs \( +y \) and \( -y \), leading to the equation \( x + x = 2x \). On the right side, you simply add the constants, resulting in \( 1 + 3 = 4 \). Now, the equation simplifies to \( 2x = 4 \), making it easy to solve for the remaining variable.

Solving Linear Equations

Solving linear equations is the process of finding the value of the variables that satisfy the equation. Once you have made use of the addition method to simplify the system, solving the equation becomes straightforward.

Let's continue with our simplified result from the addition method:

• \( 2x = 4 \)

To find \( x \), you need to isolate the variable. Divide both sides of the equation by the coefficient of \( x \), which is 2 in this case. Mathematically, this looks like:

• \( x = \frac{4}{2} = 2 \)

Once you have \( x \), plug it back into one of the original equations to solve for \( y \). For instance, substitute \( x = 2 \) back into \( x + y = 1 \):

• \( 2 + y = 1 \)
• Subtract 2 from both sides: \( y = 1 - 2 = -1 \)

The solution of the system is \( x = 2 \) and \( y = -1 \), meaning the pair \((2, -1)\) solves both original equations.

Algebraic Manipulation

Algebraic manipulation is a key skill in solving equations and involves rearranging and simplifying expressions to isolate variables and find their values. This process can involve various operations like addition, subtraction, multiplication, and division—used strategically based on problem requirements.

In the given system of equations:

• \( x + y = 1 \)
• \( x - y = 3 \)

algebraic manipulation is used at several stages. First, the equations are aligned and added, utilizing the addition method to cancel out \( y \). This simplifies the system, allowing us to handle a single equation \( 2x = 4 \).

After finding \( x \), substitution is another form of algebraic manipulation. We substitute \( x = 2 \) back into Equation 1, transforming it into a simpler equation

• \( 2 + y = 1 \)

Here, additional manipulation is needed to solve for \( y \) by isolating it on one side, resulting in the solution for the variable. Mastering these manipulations ensures that you can solve a range of algebraic problems efficiently.