Problem 16 - Algebra 1

Question

Choose a method to solve the linear system. Explain your choice. $$ \begin{aligned} &2 x-5 y=0\\\ &x-y=3 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system is \(x = 5\) and \(y = 2\). The substitution method was used for its straightforward application in this case.

Step 1: Choosing a Solution Method

Substitution method might be a good choice here as the second equation can be easily rearranged to express \(x\) or \(y\) in terms of the other variable. In this case, rearranging the second equation to express \(x\) in terms of \(y\): \(x = y + 3\).

Step 2: Substituting the Expression into the First Equation

Substitute \(x\) with \(y + 3\) in the first equation, resulting in: \(2(y + 3) - 5y = 0\).

Step 3: Solving for the Variable \(y\)

Simplify the equation to solve for \(y\). The result is: \(2y + 6 - 5y = 0 \Rightarrow -3y + 6 = 0 \Rightarrow -3y = -6 \Rightarrow y = 2\).

Step 4: Solving for the Variable \(x\)

Substitute \(y = 2\) into the rearranged second equation \(x = y + 3\) to solve for \(x\). The result is \(x = 2 + 3 = 5\).

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of linear equations by isolating one variable in terms of the other. This approach is beneficial when one of the equations is simple enough to easily solve for one of the variables.
In the given exercise, we chose the substitution method because the second equation, \( x - y = 3 \), can be straightforwardly rearranged to solve for \( x \) as \( x = y + 3 \).

• This expressiveness simplifies the process and can make the solution more manageable.
• By substituting the derived expression of one variable into the other equation, you're left with an equation that depends on a single variable, making it easier to solve.

Once one variable is found, you substitute back to find the other variable. This method is especially efficient for systems where one variable can be easily solved in terms of another.

Linear Equations

Linear equations are equations that represent straight lines when graphed on a coordinate plane. They have the general form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants.
In a system of linear equations, the intersection of these lines represents the solution to the system. Our exercise involved the following linear equations:

• \( 2x - 5y = 0 \)
• \( x - y = 3 \)

These equations together form a system of linear equations. The goal is to find a common solution \((x, y)\) that satisfies both equations simultaneously.
Such equations are fundamental in algebra as they represent an essential concept of combining multiple equations to find a united solution.

Variable Substitution

Variable substitution is the action of replacing a variable with an equivalent expression. This is at the heart of the substitution method. In our problem, once we rearranged the second equation \( x - y = 3 \) to \( x = y + 3 \), we swapped \( x \) in the first equation with \( y + 3 \), resulting in the equation \( 2(y + 3) - 5y = 0 \).
Here's why this step is crucial:

• It converts a two-variable problem into a single-variable problem, which is simpler to solve.
• After substitution, solving for \( y \) directly gives us \( y = 2 \).
• Once \( y \) is known, substituting back into \( x = y + 3 \) quickly yields \( x = 5 \).

This method makes the often daunting task of working through systems of equations more approachable by tackling one variable at a time.