Problem 15 - Precalculus with Limits

Question

Solving a System by Substitution In Exercises \(15-24\) , solve the system by the method of substitution. $$\left\\{\begin{aligned} x-y &=2 \\ 6 x-5 y &=16 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 6\) and \(y = 4\).

Step 1: Solve one equation for one variable

To isolate \(x\), we rearrange the first equation to get: \(x = y + 2\).

Step 2: Substitute the expression into the other equation

Substitute \(x = y + 2\) into the second equation: \(6(y + 2) - 5y = 16\).

Step 3: Simplify and Solve for \(y\)

Simplify and solve the equation, \(6y + 12 - 5y = 16\). This simplifies to \(y = 4\).

Step 4: Substitute \(y\) into the equation solved in Step 1 to find \(x\)

Substitute \(y = 4\) into \(x = y + 2\) to get, \(x = 4 + 2 = 6\)

Key Concepts

Substitution Method

The substitution method is a powerful technique used to find solutions for systems of linear equations. Here's how it works:

• Choose one of the equations from the system to solve for one variable in terms of the other.
• Substitute the expression obtained into the other equation.
• This substitution transforms the system into a single equation with one variable.
• Finally, solve this equation to find the value of one variable and back-substitute to find the other variable.

Let's look at our problem: you start by solving for one variable. In the system given, we took the first equation, which was simpler, and expressed \( x \) as \( x = y + 2 \). This is the key step in the substitution method, providing a straightforward path to solve the rest of the system.
Next, replace this expression into the second equation to eliminate \( x \). This allows you to focus on solving a simpler equation in just the variable \( y \). Once you have \( y \), substitute back to get \( x \). This step-by-step approach makes the substitution method ideal for solving linear systems with a manageable number of equations.

Linear Equations

Linear equations form the backbone of algebra and represent relationships where each term is either a constant or a product of a constant and a single variable. These are called linear because they graph as straight lines when plotted on a graph.In a typical scenario, a linear equation in two variables has the form \( ax + by = c \), where \( a \), \( b \), and \( c \) are constants. Their simplicity allows for clear and logical manipulation towards finding a solution. For our given problem, the equations \( x-y = 2 \) and \( 6x-5y = 16 \) are linear.Both of these are classic examples of linear equations because:

• The highest power of any variable is 1.
• All terms are either constants or products of constants and solitary variables.

These traits make linear equations incredibly versatile in modeling and solving real-world problems.

Solution of Linear Systems

To solve a system of linear equations means to find all possible values for the variables that satisfy every equation in the system simultaneously. The solution, if it exists, can be represented as an ordered pair in two dimensions.For the system given, by using the substitution method, we've derived:

• \( y = 4 \)
• Substituting back, \( x = 6 \)

Thus, the solution is \( (x, y) = (6, 4) \). This means if you substitute \( x = 6 \) and \( y = 4 \) into the original equations, both will hold true simultaneously.There are a few potential outcomes in solving linear systems:

• One unique solution, as in our example.
• No solution, which occurs if the lines are parallel.
• Infinitely many solutions, if the lines coincide entirely.

Identifying the type of solution is as important as finding the actual values because it gives insights into the relationship the equations are modeling.