Problem 14 - Algebra for College Students

Question

Solve each system by using the substitution method. \(\left(\begin{array}{l}3 x-5 y=25 \\ x=y+7\end{array}\right)\)

Step-by-Step Solution

Answer

The solution is \((5, -2)\).

Step 1: Identify the System of Equations

The given system of linear equations is: \( 3x - 5y = 25 \) and \( x = y + 7 \). These are our two equations: Equation (1) is \( 3x - 5y = 25 \) and Equation (2) is \( x = y + 7 \).

Step 2: Solve for a Variable

Notice that Equation (2), \( x = y + 7 \), already expresses \( x \) in terms of \( y \). Thus, we can directly use this expression in the substitution method.

Step 3: Substitute into the Other Equation

Substitute \( x = y + 7 \) from Equation (2) into Equation (1): \[ 3(y + 7) - 5y = 25 \]

Step 4: Simplify the Equation

Distribute and simplify the equation: \[ 3y + 21 - 5y = 25 \] Combine like terms: \[ -2y + 21 = 25 \]

Step 5: Solve for \( y \)

First, isolate the term with \( y \): \(-2y + 21 = 25\) becomes \(-2y = 25 - 21\), so \(-2y = 4\).Divide by \(-2\) to get \( y = -2 \).

Step 6: Substitute \( y \\) Back into the Other Equation

Use \( y = -2 \) in Equation (2): \[ x = y + 7 = -2 + 7 \]. Simplify this to find \( x \): \( x = 5 \).

Step 7: Write the Solution

The solution to the system of equations is \((x, y) = (5, -2)\).

Key Concepts

System of Linear Equations

A system of linear equations consists of two or more linear equations involving the same set of variables. In a typical two-variable system, these equations represent lines on a coordinate plane, and their intersection points identify possible solutions. The current exercise involves a pair of simultaneous equations:

• Equation (1): \( 3x - 5y = 25 \)
• Equation (2): \( x = y + 7 \)

Both equations include the variables \(x\) and \(y\), which we are tasked with solving.

Solving Equations

Solving equations involves finding the values of variables that make the given statement true. In this exercise, the goal is to find \(x\) and \(y\) where both equations in the system are satisfied simultaneously.
The equations are solved using mathematical methods that may include graphing, elimination, or substitution. Here, the substitution method is applied, leveraging one equation to find an expression for the variable, making it easier to solve the other equation.
Successful solving requires careful manipulation and strategic substitution to keep calculations straightforward, as demonstrated in the step-by-step solution.

Algebraic Substitution

The substitution method simplifies solving a system by replacing one variable with an expression derived from another equation. In our exercise:

• The second equation, \( x = y + 7 \), provides an expression for \(x\) in terms of \(y\).
• Substitute \(y + 7\) for \(x\) in Equation (1) results in a single equation in terms of \(y\):
  \[ 3(y + 7) - 5y = 25 \]

This substitution step is crucial, as it reduces the system to one equation with one variable. Solving it allows for an easier back-substitution to find the missing variable.

Linear Algebra

Linear algebra is the branch of mathematics concerning linear equations and their representations through matrices and vector spaces. For beginners, this often starts with solving systems of linear equations like those in the current exercise.
Understanding the process of substitution offers foundational insights into more complex concepts of linear algebra. Techniques learned here—in particular managing and transforming equations—prove valuable when dealing with larger systems of equations or matrix operations.
Though this exercise doesn't delve into matrices, the systematic approach building using substitution underscores key practices in linear algebra, such as consistency and precision in step-by-step problem-solving.