Problem 18 - Algebra 1: Concepts and Skills

Question

Choose a solution method to solve the linear system. Explain your choice, and then solve the system. $$ \begin{array}{r} {2 x+y=5} \\ {x-y=1} \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Rearrange the second equation

The second equation can be arranged to y = x - 1. It involves just moving the '-y' on the left to the right side of the equals sign.

Step 2: Substitute the second equation into the first one

Substitute y in the first equation with the expression we found in the step 1 (x - 1). So the first equation becomes 2x + x - 1 = 5. This forms a new equation in one variable x that will be simpler to solve.

Step 3: Find value of x

Solve the equation from step 2, 2x + x - 1 = 5, to find x. Firstly, combine similar terms to get 3x - 1 = 5. Add 1 to both sides, the equation becomes 3x = 6. Then divide both sides by 3 gives x = 2.

Step 4: Solve for y

Substitute x = 2 into the rearranged equation from step 1 y = x - 1, resulting y = 2 - 1= 1.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of linear equations. In simple terms, it involves expressing one variable in terms of the other. This method is useful when one equation in the system can be easily rearranged to isolate one variable. By doing so, we can simplify the computational process by focusing on one variable at a time.

Here's how it generally works:

• Choose an equation to rearrange and solve for one variable in terms of the other.
• Substitute this expression into the other equation(s) in the system.
• Solve the resulting equation to find the value of one of the variables.
• Replace this solved value back into the rearranged equation to find the other variable.

In the given exercise, the equation \( x - y = 1 \) was rearranged to \( y = x - 1 \), and this expression was substituted into the first equation \( 2x + y = 5 \). Working through substitution requires careful algebraic manipulation, which we will discuss further.

Linear Equations

Linear equations are the foundation of algebra and solving systems of linear equations is a vital skill. A linear equation is an equation of the first order; it is an equation between two variables that produces a straight line when graphed. Examples include equations like \( 2x + y = 5 \) and \( x - y = 1 \).

These equations can have one solution, no solution, or infinitely many solutions when plotted together. The goal of solving a system of linear equations is to find the values of the variables that make all equations true simultaneously.

Typically, systems of linear equations are solved using one of the following methods:

• Substitution Method
• Elimination Method
• Graphical Method

Choosing the right method depends on the equations at hand. In our case, substitution was selected due to the ease of isolating \( y \) in \( x - y = 1 \). After solving the system by substitution, the solutions \( x = 2 \) and \( y = 1 \) show where the two equations intersect.

Algebraic Manipulation

Algebraic manipulation plays a crucial role in solving equations, particularly in the context of substitution. It involves rearranging and simplifying expressions to make solving easier. The main tasks in algebraic manipulation include collecting like terms, isolating variables, and performing arithmetic operations correctly.

Let's see some common steps involved:

• Combine like terms: In step 3 of the original solution, \( 2x + x - 1 = 5 \) simplifies to \( 3x - 1 = 5 \).
• Isolate the variable: To isolate \( x \), you add 1 to both sides resulting in \( 3x = 6 \).
• Divide both sides: Finally, divide both sides by 3, giving \( x = 2 \).

Each of these steps helps untangle the puzzle of equations, allowing you to find variable values effectively. Mastery of these manipulation techniques is key to solving not just linear systems, but a whole variety of algebraic problems.