Problem 29 - Precalculus: Mathematics for Calculus

Question

Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in the ordered-pair form given in Example 6. $$\left\\{\begin{array}{c} x+2 y=7 \\ 5 x-y=2 \end{array}\right.$$

Step-by-Step Solution

Answer

The solution to the system is \((1, 3)\).

Step 1: Analyze the Given System of Equations

We have the following system of linear equations:1. \( x + 2y = 7 \)2. \( 5x - y = 2 \).Our goal is to find the values of \( x \) and \( y \) that satisfy both equations simultaneously.

Step 2: Solve for One Variable

First, let's solve the first equation for \( x \): \[ x = 7 - 2y \].This expresses \( x \) in terms of \( y \). We'll use this equation to substitute for \( x \) in the second equation.

Step 3: Substitute and Simplify

Substitute \( x = 7 - 2y \) into the second equation:\[ 5(7 - 2y) - y = 2 \].Simplify the equation to solve for \( y \):\[ 35 - 10y - y = 2 \]\[ 35 - 11y = 2 \]\[ -11y = 2 - 35 \]\[ -11y = -33 \]\[ y = 3 \].

Step 4: Substitute Back to Find x

Now that we know \( y = 3 \), substitute it back into the equation \( x = 7 - 2y \):\[ x = 7 - 2(3) \]\[ x = 7 - 6 \]\[ x = 1 \].

Step 5: Verify the Solution

Let's check the solution \((x, y) = (1, 3)\) by substituting these values back into both original equations to ensure they hold true:1. \( x + 2y = 1 + 2(3) = 1 + 6 = 7 \) (True)2. \( 5x - y = 5(1) - 3 = 5 - 3 = 2 \) (True)Both equations are satisfied, so the solution is correct.

Key Concepts

Linear Equations

Linear equations are fundamental components in algebra. They are equations featuring variables like \(x\) and \(y\), and these variables are only raised to the power of one. For instance, consider the equation \(x + 2y = 7\). The term "linear" stems from the fact that if you graphed these equations, the result would be a straight line.

Linear equations can be expressed in the form \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants. In our system, the equations can be rewritten in this standard form:
1. \(x + 2y = 7\)
2. \(5x - y = 2\)

The task is to find a common solution for these equations, which results in a particular intersection point on a graph. Solving linear equations is a critical skill, and it is often the first step in solving more complex mathematical problems.

Solving Systems

Solving systems of equations means finding values for the variables that satisfy all equations in the system simultaneously. A system of equations can have:

• A single unique solution
• Infinitely many solutions
• No solution at all

In the given exercise, we are working with two equations, forming a system of linear equations since both equations are linear. The result of solving such a system could be a specific ordered pair \((x, y)\).

When lines intersect at a point, like in our case, the ordered pair at the intersection is the solution. If the lines were parallel, there would be no solution, and identical lines would mean infinitely many solutions. Understanding the nature of the solutions helps us grasp the relationship between the lines graphically as well as algebraically.

Substitution Method

The substitution method is one of several strategies for solving systems of equations. It involves substituting one equation into the other to reduce the system to a single equation with one variable.

In our exercise, we start by isolating one variable; we take the first equation \(x + 2y = 7\) and express \(x\) in terms of \(y\) as \(x = 7 - 2y\).
This expression is then substituted into the second equation \(5x - y = 2\), transforming it into a form with just \(y\):
\(5(7 - 2y) - y = 2\).

After solving for \(y\), we backtrack to find \(x\). This method is straightforward and particularly useful when one of the equations is already solved or can be easily manipulated to solve for one variable. Mastering substitution is key to efficiently tackling diverse systems of equations in math.