Question

Solve the system by elimination Then state whether the system is consistent inconsistent. $$\left\\{\begin{aligned} 2 x-3 y &=4 \\\\-2 x-y &=4 \end{aligned}\right.$$

Short answer

The solutions are \(x = -1\) and \(y = -2\). The system of equations is consistent because there exists at least one solution that satisfies both of the equations.

Step-by-step solution

Addition of the two equations

To eliminate the variable \(x\), the two given equations are added: \[2x - 3y + (-2x - y) = 4 + 4\]. This simplifies to: \[-4y = 8\]

Solving for the variable \(y\)

By dividing both sides of the equation by -4, the value of \(y\) is found: \[y = -2\].

Substituting the value of \(y\) in one of the original equations

Substitute \(y = -2\) in the first original equation: \[2x - 3*(-2) = 4\]. This simplifies to: \[2x + 6 = 4\] or \[2x = -2\].

Solving for the variable \(x\)

By dividing both sides of the equation by 2, the value of \(x\) is found: \[x = -1\].

Verification of the solution

Substitute \(x = -1\) and \(y = -2\) in the given system of equations. They are both satisfied. So, the solution \((x,y) = (-1,-2)\) is correct.

Key concepts

Elimination Method

The Elimination Method is a popular technique for solving a system of linear equations. This method aims to eliminate one variable, making it easier to solve for the remaining variables. The basic idea is to add or subtract the equations in such a way that one of the variables cancels out.

Here’s how it works:

• First, align the equations vertically based on their terms.
• Then, manipulate the equations (by multiplying if necessary) so that one of the variables has the same coefficient in both equations.
• Next, add or subtract the equations to eliminate that variable.
• Finally, solve the resulting single-variable equation and use this solution to find the other variable.

For instance, in our given system, multiplying the equations was unnecessary as they allowed direct elimination of the variable \(x\). This makes the method efficient and effective, especially for beginners.

Consistent System

A Consistent System has at least one solution. In the context of linear equations, this means that the lines represented by the equations intersect at some point on the graph.

There are two types of consistent systems:

• Independent: Exactly one solution exists. This happens when the lines intersect at one point.
• Dependent: Infinitely many solutions exist. This happens when the lines coincide, meaning both equations essentially describe the same line.

In our example, solving by the elimination method gave us a single solution \((x, y) = (-1, -2)\). This signifies an independent consistent system. It's always useful to verify your solutions to confirm the consistency of the system by substituting the values back into the original equations.

Linear Equations

Linear equations are mathematical expressions that create straight lines when graphed. They take the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants.

Key characteristics include:

• The highest power of the variable is one, which ensures the linearity of the graph.
• The solution of a system of linear equations is the point(s) where their graphs intersect.

The simplicity of linear equations makes them foundational in algebra. In the given exercise, each equation falls perfectly into the linear equation format. The simplicity aids in using methods like elimination and substitution to find solutions effectively.

Substitution Method

The Substitution Method is another technique used to solve systems of linear equations. Here, you solve one equation for one variable and substitute that solution into the other equation.

The procedure involves:

• First, solve one of the equations for one of its variables.
• Next, substitute this expression into the other equation. This step will contain only one variable, simplifying the solution.
• Finally, solve for the remaining variable and back-substitute to find the other variable.

While our exercise focused on the elimination method, using substitution could also solve this system. For example, rearranging the second equation to express \(x\) or \(y\) in terms of the other, and substituting into the first, would eventually lead to the same solution \((x, y) = (-1, -2)\). The substitution method is often beneficial when one of the equations is easy to manipulate.