Problem 65 - Algebra 1

Question

Solve the system of linear equations if possible. Does the system have exactly one solution, no solution, or infinitely many solutions? $$ \begin{aligned}&-2 x+8 y=11\\\&x+6 y=2\end{aligned} $$

Step-by-Step Solution

Answer

The solution for the system of equations yields x = -1 and y = 1.5. The system has exactly one solution.

Step 1: Simplify the first equation

Divide the entire first equation by 2, to make it easier to solve. This will change the equation -2x + 8y = 11 into -x + 4y = 5.5

Step 2: Rearrange the first equation

Rearrange simplified first equation from Step 1 so x is on one side. This yields x = 4y - 5.5

Step 3: Substitute x from the first equation into the second

Replace x in the second equation (x + 6y = 2) with the expression we found from the first equation (4y - 5.5). This gives us 4y - 5.5 + 6y = 2.

Step 4: Solve for y from the second equation

Combine like terms and isolate y, which gives us y = 1.5.

Step 5: Substitute y into the first equation and solve for x

Plug the solved value of y into the equation from Step 2 (x = 4y -5.5) to get x = 4(1.5) - 5.5, and solve for x, which yields x = -1.

Step 6: Check the solution

Substitute the solved values of x and y into the original equations to verify if they satisfy both. If both are satisfied, then the solutions are correct.

Key Concepts

Algebraic Methods

When faced with a system of linear equations, algebraic methods offer a structured way to find solutions. These methods include substitution, elimination, and the graphical method. To understand and solve these systems, it's crucial to grasp that a solution to a system of equations represents points where the graphs of the equations intersect—meaning they have the same values for both variables.

For example, let's take our given system:

• (-2x + 8y = 11)
• (x + 6y = 2)

One must first manipulate the equations into a solvable format, which may involve simplifying or rearranging them. This approach is advantageous because it provides a step-by-step procedure to follow, reducing the complexity of finding a solution. This systematic process helps students ensure accuracy in their calculations and grasp the relationships between the variables involved.

Substitution Method

The substitution method is particularly effective for solving systems of equations where one of the variables can be isolated easily. This involves expressing one variable in terms of the other from one equation and then substituting this expression into the second equation.

Take, for instance, our exercise. Initially, one equation is simplified to (-x + 4y = 5.5), making it easier to isolate (x) as (x = 4y - 5.5). This expression for (x) is then substituted into the other equation to solve for (y). The key advantage here is that substitution reduces the system to a single equation in one variable, simplifying the solving process. This method teaches students to identify opportunities to reduce complexity and reinforce their understanding of variable relationships.

Linear Equations Solutions

The solutions to linear equations reflect the points of intersection on a graph, where each equation's line crosses. When solved correctly, these solutions disclose a lot about the system itself:

• Exactly one solution indicates that the lines intersect at a single point and are therefore non-parallel.
• No solution implies the lines are parallel and never meet.
• Infinitely many solutions mean the lines are coincident, lying on top of one another.

In our example, solving the substituted equation yields a single value for (y), and further solving for (x) reveals a single pair of values. Therefore, our system has exactly one solution, indicating the equations represent lines that intersect precisely at one point. Understanding the nature of solutions is essential for students as it enhances their comprehension of geometric interpretations of algebraic concepts.