Problem 43 - Intermediate Algebra

Question

Solve each system by any method, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} x=\frac{2}{3} y \\ y=4 x+50 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \( x = -20 \) and \( y = -30 \). The system is consistent and independent.

Step 1: Substitute Expression for x into the Second Equation

We are given the system of equations: \( x = \frac{2}{3}y \) and \( y = 4x + 50 \). Substitute the expression for \( x \) from the first equation into the second equation. This will help us express everything in terms of \( y \): \[ y = 4\left(\frac{2}{3}y\right) + 50 \].

Step 2: Simplify the Equation

Distribute the 4 through the fraction: \[ y = \frac{8}{3}y + 50 \]. We now have an equation with all terms in terms of \( y \).

Step 3: Isolate the Variable y

To isolate \( y \), subtract \( \frac{8}{3}y \) from both sides: \[ y - \frac{8}{3}y = 50 \]. Simplifying the left side, we get \[ \left(1 - \frac{8}{3}\right)y = 50 \]. This can be rewritten as \( -\frac{5}{3}y = 50 \).

Step 4: Solve for y

Continuing from the previous form \( -\frac{5}{3}y = 50 \), multiply both sides by \(-\frac{3}{5}\) to solve for \( y \): \[ y = 50 \times -\frac{3}{5} \]. Simplify the right side to find \( y = -30 \).

Step 5: Substitute y back into the First Equation

Substitute \( y = -30 \) back into the first equation to find \( x \): \[ x = \frac{2}{3}(-30) \]. Simplify to get \( x = -20 \).

Step 6: Conclusion

The solution to the system of equations is \( x = -20 \) and \( y = -30 \). This indicates the system is consistent and independent.

Key Concepts

Substitution Method

The substitution method is one of the most effective techniques for solving a system of equations. It involves solving one equation for one variable, and then using this expression to substitute into the other equation. This reduces the two-variable system into a single equation with one variable, making it easier to solve.
For instance, in our exercise, we started with the system of equations:

• Equation 1: \( x = \frac{2}{3}y \)
• Equation 2: \( y = 4x + 50 \)

We solve Equation 1 to find \( x \) in terms of \( y \), which is already done as \( x = \frac{2}{3}y \). This expression is then substituted into Equation 2, allowing us to write everything in terms of \( y \).
This method is particularly useful because it simplifies complex systems and minimizes calculations, resulting in precise solutions.

Consistent System

In the context of systems of equations, a consistent system is one which has at least one solution. This is important as it tells us that the equations in the system work together to form a valid intersection point on a graph.
In our example, after applying the substitution method, we found specific values for \( x \) and \( y \) that satisfy both equations. This indicates that our system is consistent.

• If no such solution exists, the system would be termed inconsistent.

A consistent system, like the one in our problem, often leads to practical applications since it implies that a set of criteria can all be satisfied under certain conditions. This makes it a valuable concept in various fields, including mathematics, economics, and engineering.

Solving Equations

Solving equations involves finding the values of variables that make the equation true. When dealing with systems of equations, as in our exercise, solving usually means finding the values that satisfy all equations simultaneously.
In the exercise, we used algebraic manipulation to solve for \( y \), and then substituted back to find \( x \). Let's summarize the key steps:

• Substitute the expression for \( x \) into the second equation.
• Simplify to isolate \( y \) by manipulating the equation algebraically.
• Solve for \( y \), then substitute back to find \( x \).

This process helps in understanding how changes to one variable affect another, and is pivotal in scientific and real-world problem solving.

Independent Equations

Independent equations in a system mean that each equation provides unique information and is not a scalar multiple of another. This is crucial because independent equations lead to unique solutions.
In our scenario, combining the two equations results in distinct solutions for \( x \) and \( y \), confirming that the equations are independent.

• If the equations were dependent, they would represent the same line or plane, leading to infinite solutions.
• In the exercise, we confirmed the independence by obtaining a unique solution pair \( (x, y) = (-20, -30) \).

The independence of the equations is essential as it guarantees that the solution set is not artificially enlarged, which makes the analysis straightforward and reliable.