Problem 40 - Elementary Algebra Within Reach

Question

In Exercises 35-46, solve the system by the method of substitution. $$ \left\\{\begin{array}{l} 4 x-5 y=0 \\ 2 x-5 y=-10 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = 5 \) and \( y = 4 \).

Step 1: Solve the first equation for \( x \)

Isolate \( x \) in the first equation \( 4x - 5y = 0 \). To get \( x \) by itself, add \( 5y \) to both sides and then divide by \( 4 \). This yields \( x = 1.25y \).

Step 2: Substitute expression for \( x \) into the second equation

Substitute \( x = 1.25y \) into the second equation \( 2x - 5y = -10 \). This gives \( 2(1.25y) - 5y = -10 \), which simplifies to \( 2.5y - 5y = -10 \).

Step 3: Solve for \( y \)

Solve the equation \( 2.5y - 5y = -10 \) for \( y \). Combine like terms and divide both sides by -2.5 to get \( y = 4 \).

Step 4: Substitute \( y \) into the first equation

Substitute \( y = 4 \) into the first equation to find \( x \). This gives \( 4x - 5(4) = 0 \) or \( 4x = 20 \), which simplifies to \( x = 5 \).

Key Concepts

Solving Systems of Equations

When we talk about solving systems of equations, we are referring to finding the values of the variables that satisfy all the equations in the system simultaneously. Typically, a system of linear equations can have one solution, no solution, or infinitely many solutions. In this exercise, we have two linear equations:

• \(4x - 5y = 0\)
• \(2x - 5y = -10\)

The goal is to determine the values of \(x\) and \(y\) that make both equations true. Solving such a system involves finding the intersection point of the lines represented by the equations on a graph. However, instead of graphing, we can use algebraic methods, such as substitution or elimination.

Linear Equations

Linear equations in two variables, like \(4x - 5y = 0\) and \(2x - 5y = -10\), represent straight lines when plotted on a graph. The term 'linear' signifies that the graph of the equation forms a straight line, and each equation can be written in the form \(ax + by = c\). In our example, both equations have the same form with different constants.

When dealing with linear equations, it's crucial to understand their structure:

• The coefficients \(a\) and \(b\) indicate the steepness or slope of the line.
• The constant \(c\) determines the line's position on the graph.

These properties help us comprehend how the equations will interact and at which point they will intersect or if they are parallel. Here, since the system is consistent and independent, its solution is a single intersection point, meaning the lines meet at only one point.

Substitution Method

The substitution method is a popular and effective technique for solving a system of equations, particularly when it involves linear equations. This method involves expressing one variable in terms of the other and then substituting this expression into the second equation. Here's how it works in this example:

• First, solve one equation for one variable. In the exercise, we solved the first equation, \(4x - 5y = 0\), for \(x\), yielding \(x = 1.25y\).
• Next, substitute \(x = 1.25y\) into the second equation \(2x - 5y = -10\). This replaces the \(x\) in the second equation, allowing us to solve for \(y\).
• After finding the value of \(y\), substitute it back into one of the original equations to find \(x\). In this case, substituting \(y = 4\) back gives us \(x = 5\).

By following these steps, we solve the original system entirely algebraically, without the need to graph the equations. This method is especially useful when one equation is easily solvable for one variable.