Solving Dependent Linear Systems with Substitution

In the world of algebra, systems of equations are like puzzles waiting to be solved. We’re looking for the values of variables that make all equations in the system true. One way to tackle these puzzles is using the substitution method. Today, we’ll explore how to solve a special type of system: a **dependent linear system**.

What are Dependent Linear Systems?

Imagine two lines on a graph. A dependent system is where these lines are identical, meaning they overlap perfectly. This means there are infinitely many points where the lines intersect, representing infinitely many solutions to the system.

Here’s the key to recognizing a dependent system:

• **The equations are multiples of each other:** One equation can be obtained by multiplying the other equation by a constant.
• **When you try to solve the system, you end up with a true statement, but no unique solutions:** You’ll see this as a statement like 0 = 0.

Solving with Substitution: Step-by-Step

Let’s dive into how to solve a dependent linear system using substitution. Here’s a breakdown of the steps:

1. Solve one equation for one variable: Choose one equation and solve it for one of the variables. It doesn’t matter which one you pick, but choose the equation that makes it easiest to isolate a variable.
2. Substitute the expression into the other equation: Take the expression you just found and substitute it into the other equation. This will create an equation with only one variable.
3. Solve for the remaining variable: Solve the new equation for the remaining variable. You’ll likely end up with a true statement like 0 = 0.
4. Express the solution in terms of a parameter: Since there are infinitely many solutions, we’ll use a parameter (often represented by ‘t’) to describe the solution set.
5. Write the solution set: Express the solution in the form of an ordered pair (x, y) using the parameter ‘t’.

Example: Let’s Solve!

Let’s work through an example to see this in action. Consider the following system:

Equation 1: 2x + 3y = 6

Equation 2: 4x + 6y = 12

1. Solve for x in Equation 1:
   2x = 6 – 3y
   x = 3 – (3/2)y
2. Substitute x into Equation 2:
   4(3 – (3/2)y) + 6y = 12
   12 – 6y + 6y = 12
   12 = 12
3. Solve for y: We get a true statement (12 = 12), indicating infinitely many solutions.
4. Express the solution in terms of a parameter: Let y = t (where ‘t’ is any real number).
5. Write the solution set:
   x = 3 – (3/2)t
   y = t
   Solution: {(3 – (3/2)t, t) | t ∈ ℝ}

This solution tells us that for any value of ‘t’, we can find a corresponding value for x and y that satisfies both equations.

Key Points to Remember

• Dependent linear systems represent overlapping lines with infinitely many solutions.
• Substitution method can be used to solve them, leading to a true statement and no unique solutions.
• The solution set is expressed using a parameter ‘t’ to represent all possible solutions.

Understanding dependent linear systems and how to solve them using substitution is a valuable skill in algebra. It helps you navigate the world of equations and interpret the relationships between lines on a graph.