Problem 17 - Algebra 1

Question

Choose a method to solve the linear system. Explain your choice. $$ \begin{aligned} &3 x+2 y=10\\\ &2 x+5 y=3 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x = 4, y = -1\)

Step 1: Multiplication

To prepare for the elimination method, you could make the coefficients of one of the variables the same in both equations. In this case, we can multiply the first equation by 2 and the second equation by 3. The system becomes: \[ \begin{aligned} 6x + 4y = 20\ 6x + 15y = 9\ \end{aligned} \]

Step 2: Elimination

Subtract the second equation from the first to eliminate 'x'. The result is: \(-11y = 11\), which simplifies to \(y = -1\).

Step 3: Solving for the other variable

Now substitute \(y = -1\) into either of the original equations. Using the first original equation (3x + 2y = 10), the equation becomes \(3x + 2*(-1) = 10\), which simplifies to \(3x = 12\) and \(x = 4\).

Key Concepts

Elimination Method

When solving systems of linear equations, the elimination method is a powerful tool to understand. It involves manipulating the equations to eliminate one of the variables, making it easier to solve for the other. Consider the system of equations from the problem:
\[\begin{aligned}&3x + 2y = 10 \&2x + 5y = 3\end{aligned}\]
The key step in using the elimination method is to adjust the coefficients of one variable so they are the same in both equations. This is accomplished by multiplying the entire equation by a suitable number:

• Multiply the first equation by 2.
• Multiply the second equation by 3.

This changes the system of equations to:
\[\begin{aligned}6x + 4y = 20 \6x + 15y = 9\end{aligned}\]
Now, subtract the second equation from the first to eliminate the variable \(x\):
\(-11y = 11\) simplifies to \(y = -1\). With \(y\) found, you can use substitution to find \(x\). The elimination method is particularly useful when dealing with systems where one can conveniently cancel a variable. It simplifies calculations and can be more straightforward than substitution in some cases.

Substitution

The substitution method is another essential technique for solving systems of linear equations. Unlike elimination, substitution involves solving one of the equations for one variable and then substituting this expression into the other equation. Although it wasn't used as the primary method in the solution provided, it played a crucial role in finding the second variable after elimination was used.
Let's see how substitution can be applied once a variable, like \(y\), is known. After using the elimination method to determine that \(y = -1\), you can substitute \(y = -1\) back into either of the original equations to solve for \(x\).
Choose the first equation:
\[3x + 2(-1) = 10\]
This simplifies to:
\[3x - 2 = 10\]
Adding 2 to both sides gives:
\[3x = 12\]
Dividing by 3, we get:
\[x = 4\]

• This shows how substitution aids in finding the value of the remaining variable once one is known.

The substitution method is often more effective when one equation is easily solved in terms of one variable, leading to simpler substitution into the other equation.

Linear Equations

Linear equations represent straight lines on a graph and are fundamental components of linear algebra. Each linear equation is generally in the form \(ax + by = c\), representing a relationship between \(x\) and \(y\). In the provided exercise, both equations in the system are linear:
\[\begin{aligned}&3x + 2y = 10 \&2x + 5y = 3\end{aligned}\]
Key characteristics of linear equations include:

• They graph as straight lines, and a system of linear equations can be represented by multiple intersecting lines.
• The solution to a system of linear equations represents the point or points where the lines intersect.
• If two lines intersect at a single point, as seen in this example after solving the system, there is one unique solution.

Understanding linear equations is crucial since many real-world relationships can be expressed this way. The simplicity of linear equations makes them a starting point for exploring more complex systems of algebraic expressions. They lay the groundwork for understanding concepts like slopes, intercepts, and systems of equations.