Question
Which of the following systems of equations would best be solved using the elimination
method?
Select the correct two systems of equations.
y=4z+1
4z-3y = 18
6z +3y =102
z+5y=6
-z+7y=6
z=l1y+8
4z-15y = 90
method?
Select the correct two systems of equations.
y=4z+1
4z-3y = 18
6z +3y =102
z+5y=6
-z+7y=6
z=l1y+8
4z-15y = 90
Answers
Answered by
GPT-4o mini
To determine which systems of equations can be best solved using the elimination method, we need to look for systems where the elimination method can simplify finding the solution through addition or subtraction of the equations to eliminate one of the variables.
Let's examine the provided systems:
1)
\[
\begin{align*}
y &= 4z + 1 \quad (1) \\
4z - 3y &= 18 \quad (2)
\end{align*}
\]
2)
\[
\begin{align*}
6z + 3y &= 102 \quad (1) \\
z + 5y &= 6 \quad (2)
\end{align*}
\]
3)
\[
\begin{align*}
-z + 7y &= 6 \quad (1) \\
z = l1y + 8 \quad (2)
\end{align*}
\]
4)
\[
\begin{align*}
4z - 15y &= 90 \quad (1) \\
z + 5y &= 6 \quad (2)
\end{align*}
\]
### Analysis
**System 1:**
- The first equation can be rearranged to express \(y\) in terms of \(z\), which could then be substituted into the second equation. Not particularly suited for elimination as it relies on substitution.
**System 2:**
- Both equations are in standard form (Ax + By = C) but they do not readily lend themselves to direct elimination without modification (e.g., multiplying to align coefficients).
**System 3:**
- The second equation has a term \(z\) isolated, but like system 1, is better suited for substitution rather than elimination since one equation is not in the standard linear format.
**System 4:**
- The equations here are both in Ax + By = C format which is directly suited for the elimination method. You can eliminate \(z\) or \(y\) relatively easily by multiplying and adding or subtracting these equations.
### Conclusion
The best candidates for using the elimination method based on the formats of the equations are:
- **System 2:** \(6z + 3y = 102\) and \(z + 5y = 6\)
- **System 4:** \(4z - 15y = 90\) and \(z + 5y = 6\)
So the two systems of equations that would best be solved using the elimination method are:
**System 2 and System 4.**
Let's examine the provided systems:
1)
\[
\begin{align*}
y &= 4z + 1 \quad (1) \\
4z - 3y &= 18 \quad (2)
\end{align*}
\]
2)
\[
\begin{align*}
6z + 3y &= 102 \quad (1) \\
z + 5y &= 6 \quad (2)
\end{align*}
\]
3)
\[
\begin{align*}
-z + 7y &= 6 \quad (1) \\
z = l1y + 8 \quad (2)
\end{align*}
\]
4)
\[
\begin{align*}
4z - 15y &= 90 \quad (1) \\
z + 5y &= 6 \quad (2)
\end{align*}
\]
### Analysis
**System 1:**
- The first equation can be rearranged to express \(y\) in terms of \(z\), which could then be substituted into the second equation. Not particularly suited for elimination as it relies on substitution.
**System 2:**
- Both equations are in standard form (Ax + By = C) but they do not readily lend themselves to direct elimination without modification (e.g., multiplying to align coefficients).
**System 3:**
- The second equation has a term \(z\) isolated, but like system 1, is better suited for substitution rather than elimination since one equation is not in the standard linear format.
**System 4:**
- The equations here are both in Ax + By = C format which is directly suited for the elimination method. You can eliminate \(z\) or \(y\) relatively easily by multiplying and adding or subtracting these equations.
### Conclusion
The best candidates for using the elimination method based on the formats of the equations are:
- **System 2:** \(6z + 3y = 102\) and \(z + 5y = 6\)
- **System 4:** \(4z - 15y = 90\) and \(z + 5y = 6\)
So the two systems of equations that would best be solved using the elimination method are:
**System 2 and System 4.**
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