Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Solve the system by the method of substitution. $$\left\\{\begin{array}{r}-\frac{2}{3} x+y=2 \\ 2 x-3 y=6\end{array}\right.$$

Step-by-Step Solution

Answer

The system of equations does not have a solution. This is due to the two lines being parallel and not intersecting.

Step 1: Isolate a variable

Choose the first equation to isolate the y-variable. First, multiply entire equation by 3 to get rid of fractions: \( -2x + 3y = 6 \). Then add 2x to both sides to isolate y: \( 3y = 2x + 6 \). Lastly, divide by 3 to solve for y: \( y = \frac{2x}{3} + 2 \)

Step 2: Substitute into second equation

Substitute the equation for y into the second equation: \( 2x - 3(\frac{2x}{3} + 2) = 6 \). Simplify the equation by using the distributive property and combining like terms: \( 2x - 2x - 6 = 6 \). This further simplifies to: \( -6 = 6 \).

Step 3: Contradiction and conclusion

-6 cannot equal 6, so the system of equations does not have a solution. This happens when the two lines represented by the equations are parallel and do not intersect.

Key Concepts

Algebraic Method

The algebraic method is a foundational aspect of solving systems of equations. It involves manipulating equations to isolate variables and substitute them into other equations to find solutions. To start, isolate one variable in one of the equations. In the given exercise, we isolate the variable y in the first equation. We do this by getting rid of any fractions and moving terms around. It’s essential to maintain the equality by performing the same operations on both sides of the equation.

Once y is isolated, as in the solved exercise \( y = \frac{2x}{3} + 2 \), we can substitute this expression for y into the other equation. This allows for one equation with one unknown, x, which simplifies the problem significantly. The algebraic method is all about these small, logical steps that get us closer to the solution.

Solving System of Equations

When we talk about solving a system of equations, we mean finding the set of values for the variables that satisfy all equations simultaneously. There are several methods to achieve this, with substitution being one of them. After isolating one variable, substituting it into another equation should ideally lead to a solvable equation for the remaining variable.

If the resulting equation simplifies to a true statement, like \(0=0\), it indicates that the system has infinite solutions. However, if it simplifies to a false statement like \( -6 = 6 \), then there is no solution. It's crucial to pay attention to these nuances as they reveal the nature of the relation between the equations in the system.

No Solution Systems

A system with no solution, also known as an inconsistent system, means that there are no points of intersection between the lines represented by the equations. This situation arises when we derive a false statement during the solving process, indicating that the equations contradict each other.

For example, the exercise given presents a contradiction \( -6 = 6 \), signaling that no solution exists. In real-world situations, this could represent two conflicting conditions that cannot be met simultaneously. Recognizing no-solution systems is an important skill, as it can save time from seeking a solution that does not exist.

Parallel Lines

Parallel lines have the same slope but different y-intercepts, which means they will never intersect. In the context of algebra, the equations of parallel lines result in a system with no solution. When we simplify the substituted equation and come up with a contradiction, this is indicative of parallel lines. It is the graphical version of the algebraic contradiction.

Understanding parallel lines in algebra is crucial for visualizing why certain systems have no solutions. Identifying parallelism can often be done by analyzing the standard form coefficients of the lines (Ax + By = C) and recognizing equal ratios between the A and B coefficients, with varying C values.