Problem 45 - Algebra 1

Question

Solve the linear system. $$ \begin{aligned} &\frac{1}{2} x-y=-5\\\ &x-\frac{1}{3} y=0 \end{aligned} $$

Step-by-Step Solution

Answer

The solution to the system of equations is \(x=10, y=30\).

Step 1: Rearrange the second equation

Rearrange the second equation \(x-\frac{1}{3} y=0\) to solve for \(x\). This gives \(x=\frac{1}{3} y\).

Step 2: Substitute \(x\) into the first equation

Now that we know \(x=\frac{1}{3} y\), substitute this value into the first equation. This gives \(\frac{1}{2} (\frac{1}{3} y)-y=-5\) which simplifies to \(-\frac{1}{6} y=-5\).

Step 3: Solve for \(y\)

Now solve for \(y\) in the equation \(-\frac{1}{6} y=-5\) to get \(y=30\).

Step 4: Substitute \(y\) into the first equation to solve for \(x\)

Substitute \(y=30\) into the equation \(x=\frac{1}{3}y\) to evaluate \(x\). This gives \(x=\frac{1}{3}*30=10\).

Key Concepts

Linear Equations

Linear equations are fundamental mathematical expressions that form straight lines when graphed on a coordinate plane. They typically take the general form of \(ax + by = c\), where \(a\), \(b\), and \(c\) are constants, and \(x\) and \(y\) are variables. In this specific exercise, we are dealing with two linear equations:

• \(\frac{1}{2}x - y = -5\)
• \(x - \frac{1}{3}y = 0\)

These equations represent lines in a 2-dimensional space. The solution to this system is the point where these two lines intersect. This intersection point gives us the values of \(x\) and \(y\) that satisfy both equations simultaneously. Understanding linear equations is crucial because they are widely used to model real-world problems where variables are linearly related.

Substitution Method

The substitution method is a technique used to solve systems of equations, particularly linear systems, by expressing one variable in terms of the other. Here's a brief guide on how substitution works:

• Select one of the equations and solve for one variable.
• Substitute that expression into the other equation.
• Solve the new equation for the remaining variable.
• Use the value of that variable to find the value of the other variable.

In our specific problem:

• We started by rearranging the second equation to express \(x\) as \(x = \frac{1}{3}y\).
• Then, substituted \(x = \frac{1}{3}y\) into the first equation \(\frac{1}{2}x - y = -5\). This substitution leads to a single equation in terms of \(y\).
• Finally, we solved this equation to find the value of \(y\).

The substitution method is particularly useful when one of the equations is easy to rearrange, making it a simple but effective way to find solutions to a system.

Algebraic Manipulation

Algebraic manipulation involves rearranging and simplifying equations to solve for unknown variables. It's a critical skill in solving systems of linear equations. Let's break down the steps:

• **Rearranging Equations:** We rearranged one of our equations, \(x - \frac{1}{3}y = 0\), to isolate \(x\). This gave us \(x = \frac{1}{3}y\), making it easier to substitute into the other equation.
• **Substitution and Simplification:** Once substituted, we simplified \(\frac{1}{2}(\frac{1}{3}y) - y = -5\) to a simpler form \(-\frac{1}{6}y = -5\).
• **Solving for Variables:** With the simplified expression \(-\frac{1}{6}y = -5\), we multiplied both sides by \(-6\) to isolate \(y\), resulting in \(y = 30\). After finding \(y\), we substituted it back to find \(x = 10\).

Algebraic manipulation offers a robust framework to systematically handle and resolve the equations, ensuring clarity and correctness in the results obtained.