Problem 62 - Elementary Algebra Within Reach

Question

In Exercises 59-62, solve the system by the method of substitution. $$ \left\\{\begin{array}{r} 5 x+6 y=21 \\ 25 x+30 y=10 \end{array}\right. $$

Step-by-Step Solution

Answer

This system of equations has no solution.

Step 1: Rearrange one equation

Firstly, rearrange the first equation to solve for one variable. Perhaps solving for \(x\) in the first equation will make it easier. After rearranging, the first equation is therefore \(x = (21 - 6y) /5\).

Step 2: Substitute into the second equation

Substitute \(x\) from the first equation into the second equation: \(25(21 - 6y)/5 + 30y = 10\). Simplify the equation: \(105 - 30y + 30y = 10\). Further simplifying, this results in \(105 = 10\), which appears to be a contradiction.

Step 3: Conclude

Since no values of \(y\) would make the second equation true once substituted, it can be concluded that this system of equations has no solution. It suggests that these two lines are parallel in the Cartesian plane and hence, do not intersect.

Key Concepts

Substitution Method

The substitution method is one of the most popular techniques for solving systems of equations. It involves solving one of the equations for one variable in terms of the other. This expression is then substituted into the second equation.

• Initially, this allows you to solve for one variable.
• After calculating this variable, you can find the other by plugging back into one of the original equations.

The power of substitution lies in simplifying a problem with two unknowns into a single-variable equation, which is often much easier to solve. In this exercise, we first rearranged the first equation to express \( x \) in terms of \( y \), setting the stage for effective substitution.

Algebraic Solution

An algebraic solution involves using algebra techniques to manipulate equations. With the substitution method, the focus is on solving equations step by step, ensuring accuracy and consistency. For this system:- We first solve for \( x \) in terms of \( y \) from one given equation: \( x = \frac{21 - 6y}{5} \).- Substitute this expression for \( x \) into the other equation to find possible \( y \) values.This transforms a two-variable system into a single equation, making complex solutions more manageable. By working through the exercise, you learn an efficient way to apply algebra to solve systems using substitution.

Parallel Lines

In the context of solving systems of equations, the concept of parallel lines emerges when two equations describe lines that never meet. Lines are parallel if their slopes are equal, implying they won't intersect at any point on the Cartesian plane.

• In our exercise, simplifying both equations leads to a contradiction when substituting values.
• When solving, you found that both equations dependent on constants confirm the parallel nature directly.

Understanding parallel lines in systems of equations highlights that some systems might lack a common solution due to the alignment of their graphical representations.

No Solution Concept

A key outcome when analyzing systems of equations is the no solution scenario, meaning these systems have no intersection point on their graphs.- This arises naturally from the exercise when solving the modified equations leads to a contradiction, like \( 105 = 10 \). - A contradiction indicates the system of equations is inconsistent and has no solution. In practical terms, the two lines don't meet, illustrating that they're parallel. A clear grasp of no solution concepts strengthens your understanding of system dynamics beyond simple numerical computation. Recognizing these situations provides deep insights into the nature of systems and their graphical interactions.