Problem 11 - Algebra and Trigonometry

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 3. $$\left\\{\begin{aligned} x+3 y=5 \\ 2 x-y=3\end{aligned}\right.$$

Step-by-Step Solution

Answer

The system has a unique solution: \((2, 1)\).

Step 1: Write the System of Equations

We have two linear equations in the system, which we need to solve. The system of equations is given by: \[ \begin{align*} x + 3y &= 5 \ 2x - y &= 3 \end{align*} \]

Step 2: Solve Equation for x

Start by isolating \(x\) in the first equation. We rewrite \(x + 3y = 5\) as \(x = 5 - 3y\).

Step 3: Substitute for x

Substitute \(x = 5 - 3y\) into the second equation: \(2(5 - 3y) - y = 3\).

Step 4: Simplify the Equation

Distribute the 2: \( 10 - 6y - y = 3\), which simplifies to \(10 - 7y = 3\).

Step 5: Solve for y

Rearrange the equation to solve for \(y\): \(-7y = 3 - 10\) leading to \(-7y = -7\). Upon dividing both sides by -7, we get \(y = 1\).

Step 6: Substitute Back to Find x

Use \(y = 1\) in the equation \(x = 5 - 3y\) to find \(x\). Substitute \(y = 1\) into \(x = 5 - 3y\) resulting in \(x = 5 - 3(1) = 5 - 3 = 2\).

Step 7: Solution Verification

Check the solution by substituting \(x = 2\) and \(y = 1\) into the original equations to ensure they hold true: \(x + 3y = 5\) becomes \(2 + 3(1) = 5\) which is true, and \(2x - y = 3\) becomes \(2(2) - 1 = 3\) which is also true.

Key Concepts

Linear Equations

Linear equations are equations that involve only first-degree variables, such as \(x\) or \(y\), without any exponents. They form a straight line when graphed on a coordinate plane. In the context of the problem, we have two linear equations: one is \(x + 3y = 5\) and the other is \(2x - y = 3\).
These equations are called "linear" because their solutions can be graphed as lines on a two-dimensional plane. Linear equations can have one variable or more, but in systems of linear equations, we often deal with two or more variables.

• Important terms: Variables, coefficients, constant terms.
• Graph characteristic: A straight line on the graph.

Understanding linear equations is crucial as they serve as the foundation for linear algebra and many applications in science and engineering.

Solving Linear Systems

When we deal with linear systems, we work with multiple linear equations at the same time. Our goal is to find values for the variables that satisfy all equations simultaneously. In this exercise, our system is made up of two linear equations.
To solve a system such as \(x + 3y = 5\) and \(2x - y = 3\), we need a method to find the common solution for both equations. This common solution means we're searching for a specific \(x\) and \(y\) that simultaneously satisfy both equations.

• Types of solutions: One solution, no solution, or infinitely many solutions.
• Approaches: Graphing, substitution, elimination.

In practical terms, solving linear systems is about finding points where the lines intersect, which is why intersecting lines usually have a unique solution.

Substitution Method

The substitution method is a commonly used strategy for solving systems of linear equations. It involves expressing one variable in terms of another and substituting this expression into the other equation. This simplifies the system to one equation with one variable.
In this exercise, we isolated \(x\) in the first equation to get \(x = 5 - 3y\), and then substituted this into the second equation, transforming it into a single-variable equation.

• Steps involved: Solve one equation for a variable, substitute in the other, solve for one variable, and back-solve for the second.
• Benefits: Directly reduces the system to one equation.

This method is particularly effective when one of the equations is easy to rearrange. Once we find the value for one variable, we substitute back into one of the original equations to find the other. This provides a complete solution to the system.