Problem 47 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 0.5 x+0.5 y=6 \\ \frac{x}{2}-\frac{y}{2}=-2 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(x = 4\) and \(y = 8\).

Step 1: Identify the system of equations

The given system of equations is \(0.5x + 0.5y = 6\) and \(\frac{x}{2} - \frac{y}{2} = -2\). We need to determine if they are consistent or dependent, and solve for \(x\) and \(y\) if possible.

Step 2: Convert fractions to decimals

The equation \(\frac{x}{2} - \frac{y}{2} = -2\) can be rewritten in decimal form as \(0.5x - 0.5y = -2\) to make it easier for calculation.

Step 3: Add the equations

Add \(0.5x + 0.5y = 6\) and \(0.5x - 0.5y = -2\) to eliminate \(y\): \[ (0.5x + 0.5y) + (0.5x - 0.5y) = 6 + (-2) \]The resulting equation is \(1.0x = 4\).

Step 4: Solve for x

Divide both sides of the equation \(1.0x = 4\) by 1.0 to solve for \(x\), giving \(x = 4\).

Step 5: Substitute x into the first equation

Substitute \(x = 4\) into the first equation \(0.5x + 0.5y = 6\):\[0.5(4) + 0.5y = 6\]\[2 + 0.5y = 6\].

Step 6: Solve for y

Subtract 2 from both sides:\[0.5y = 4\]Divide both sides by 0.5:\[y = 8\].

Step 7: State the solution

The solution to the system of equations is \(x = 4\) and \(y = 8\) which means the system is consistent.

Key Concepts

Consistent Systems

In the context of systems of linear equations, a consistent system is one where there is at least one set of values for the variables that satisfies all the equations simultaneously. In other words, the lines represented by these equations intersect at at least one point on a graph.
For example, in the original exercise, the system given is \(0.5x + 0.5y = 6\) combined with \(0.5x - 0.5y = -2\). By solving this system, we arrived at a specific solution, \(x = 4\) and \(y = 8\). This confirms that the system is consistent because these values satisfy both equations. Therefore, the system is solvable, and the lines intersect at one particular point.

Dependent Equations

Dependent equations are a subset of systems where all the equations describe the same geometric line, resulting in an infinite number of solutions. This means you can find numerous combinations for \(x\) and \(y\) that satisfy the equations.

• These lines are parallel and coincide with each other, lying perfectly one on top of the other.
• If two equations in a system are dependent, solving one essentially solves the other.

In our exercise, the system wasn't dependent because solving the equations led us to a unique solution instead of infinitely many. Hence, the two equations describe two different lines that meet at a single intersection point.

Solving Linear Equations

Solving linear equations involves finding the values of variables that convert the equation into a true statement. There are several methods to achieve this:

• Substitution Method: Substitute the expression for one variable into the other equation.
• Elimination Method: Add or subtract equations to eliminate one of the variables.
• Graphical Method: Graph each equation and look for points of intersection.

In our example, we used the elimination method to simplify the equations. By adding \(0.5x + 0.5y = 6\) and \(0.5x - 0.5y = -2\), we strategically canceled out \(y\), making it easier to solve directly for \(x\). This approach can quickly lead to a solution without needing to manipulate the equations too much.

Algebraic Methods

Algebraic methods refer to techniques used in algebra to solve equations systematically and logically. The choice of method depends on the structure of the equations and the desired clarity in solving.

• Conversion to Decimals: It simplifies the arithmetic, as was done when \( \frac{x}{2} - \frac{y}{2} = -2 \) became \( 0.5x - 0.5y = -2 \).
• Linear Combination: Methods like elimination rely heavily on combining equations to simplify the problem.
• Direct Calculation: Solving simple equations through basic operations like addition, subtraction, multiplication, or division.

Each of these strategies is an algebraic tool that students can use to tackle complex linear systems progressively. In our exercise, using decimals made arithmetic operations straightforward while the elimination approach allowed focused isolation of one variable, streamlining the problem-solving process.