5x - Y = 6x + 6y = 4To Solve The System By Elimination, What Integer Would You Multiply The FIRST Equation
When dealing with systems of equations, especially those involving multiple variables, the elimination method is a powerful technique to find solutions efficiently. The question often arises: "What integer should I multiply the first equation by to eliminate one variable?" In this article, we'll delve into this specific problem involving the system:
\[
5x - y = 6x + 6y = 4
\]
and explore how to determine the appropriate integer to multiply the first equation to facilitate elimination. We will walk through the steps, clarify the reasoning, and provide tips for solving similar systems.
---
Understanding the System of Equations
Before jumping into the solution, it's essential to interpret the given system correctly. The expression:
\[
5x - y = 6x + 6y = 4
\]
can be confusing if read as a single statement. Typically, in systems of equations, each equation is separated by an equality sign. The notation here suggests that the entire expression might be a typo or a misinterpretation.
Possible interpretations:
- Two separate equations:
\[
\begin{cases}
5x - y = 4 \\
6x + 6y = 4
\end{cases}
\]
- A combined statement with a typo: Maybe it was meant as:
\[
5x - y = 6x + 6y = 4
\]
which doesn't make sense because a chain of equalities like that isn't valid unless it signifies multiple equalities.
Most likely, the intended system is:
\[
\begin{cases}
5x - y = 4 \quad (Equation\,1) \\
6x + 6y = 4 \quad (Equation\,2)
\end{cases}
\]
We will proceed under this assumption, as it aligns with common systems of two equations with two variables.
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Setting Up the Equations for Elimination
Given the two equations:
\[
\begin{cases}
5x - y = 4 \quad (Equation\,1) \\
6x + 6y = 4 \quad (Equation\,2)
\end{cases}
\]
The goal of elimination is to eliminate one variable by making the coefficients of that variable opposites in both equations.
Step 1: Identify the coefficients of the variable to eliminate.
- For \( y \): coefficients are \(-1\) in Equation 1 and \(6\) in Equation 2.
- For \( x \): coefficients are \(5\) in Equation 1 and \(6\) in Equation 2.
Suppose we choose to eliminate \( y \).
Step 2: Find the least common multiple (LCM) of the coefficients of \( y \):
\[
\text{LCM of } 1 \text{ and } 6 = 6
\]
Step 3: Determine multipliers to make the coefficients of \( y \) opposites.
- Multiply Equation 1 by 6:
\[
6 \times (5x - y) = 6 \times 4 \Rightarrow 30x - 6y = 24
\]
- Multiply Equation 2 by 1:
\[
1 \times (6x + 6y) = 4 \Rightarrow 6x + 6y = 4
\]
Now, the coefficients of \( y \) are \(-6\) and \(6\), which are opposites, allowing elimination by addition.
---
Determining the Integer to Multiply the First Equation
In our previous step, we multiplied Equation 1 by 6 to match the coefficient of \( y \):
\[
6 \times \text{Equation 1}
\]
This multiplication ensures the coefficients of \( y \) are \(-6\) and \(6\), enabling the elimination process.
Answer: The integer you multiply the first equation by is 6.
This is the minimal positive integer that facilitates eliminating \( y \) from the system.
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Performing the Elimination Method
Let's proceed with the elimination step-by-step.
Step 1: Write the modified equations:
\[
\begin{cases}
30x - 6y = 24 \quad (Equation\,1\,multiplied) \\
6x + 6y = 4 \quad (Equation\,2\,unchanged)
\end{cases}
\]
Step 2: Add the equations to eliminate \( y \):
\[
(30x - 6y) + (6x + 6y) = 24 + 4
\]
\[
(30x + 6x) + (-6y + 6y) = 28
\]
\[
36x = 28
\]
Step 3: Solve for \( x \):
\[
x = \frac{28}{36} = \frac{7}{9}
\]
Step 4: Substitute \( x = \frac{7}{9} \) back into one of the original equations to find \( y \). Using Equation 1:
\[
5x - y = 4
\]
\[
5 \times \frac{7}{9} - y = 4
\]
\[
\frac{35}{9} - y = 4
\]
\[
-y = 4 - \frac{35}{9}
\]
Express 4 as \(\frac{36}{9}\):
\[
-y = \frac{36}{9} - \frac{35}{9} = \frac{1}{9}
\]
\[
y = -\frac{1}{9}
\]
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Final Solution to the System
The solution to the system, considering our calculations, is:
\[
\boxed{
x = \frac{7}{9}, \quad y = -\frac{1}{9}
}
\]
This pair \(\left(\frac{7}{9}, -\frac{1}{9}\right)\) satisfies both original equations.
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Summary: What Integer to Multiply the First Equation?
Key Takeaways:
- To eliminate a variable, multiply the equations so that the coefficients of that variable are opposites.
- In our example, multiplying the first equation by 6 achieved this goal for \( y \).
- The choice of multiplier depends on the coefficients of the variable to eliminate; find the least common multiple (LCM) to keep calculations straightforward.
- Always verify your solution by substituting back into the original equations.
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Additional Tips for Solving Systems by Elimination
- Identify the variable to eliminate: Choose the variable with coefficients that are easiest to manipulate.
- Find the least common multiple (LCM): Use it to determine the multipliers for each equation.
- Multiply each equation appropriately: Ensure the coefficients of the chosen variable are negatives of each other.
- Add or subtract equations: This will eliminate the chosen variable.
- Solve for the remaining variable: Once one variable is eliminated, solve the resulting equation.
- Back-substitute: Substitute the found value into one of the original equations to find the other variable.
- Check your solution: Always verify by plugging your solutions into the original equations.
Conclusion
In solving systems of equations via the elimination method, choosing the correct integer multiplier for the first equation is crucial for simplifying the process. As demonstrated, multiplying the first equation by 6 allowed us to eliminate \( y \) efficiently and solve for \( x \) and \( y \). The key is analyzing the coefficients, finding the LCM, and applying the multipliers accordingly. With practice, identifying these multipliers becomes intuitive, making the elimination method a reliable tool for solving various systems of equations.
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Remember: The goal is to manipulate equations to create matching (or opposite) coefficients for one variable, enabling straightforward elimination and solution derivation. Mastering this technique will enhance your algebra skills and prepare you for more complex mathematical problems.
