Asked by Scar
What is the solution to the system of equations y = 6x - 1 and y = 3x + 2? Complete the table to support your answer.
(6 points)
x values y = 6x - 1 y = 3x + 2
-1
0
1
2
3
The ordered pair that is a solution to the system of equations is
(6 points)
x values y = 6x - 1 y = 3x + 2
-1
0
1
2
3
The ordered pair that is a solution to the system of equations is
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To find the solution to the system of equations \(y = 6x - 1\) and \(y = 3x + 2\), we can start by equating the two equations since they both equal \(y\):
\[
6x - 1 = 3x + 2
\]
Next, we solve for \(x\):
1. Subtract \(3x\) from both sides:
\[
6x - 3x - 1 = 2
\]
This simplifies to:
\[
3x - 1 = 2
\]
2. Add \(1\) to both sides:
\[
3x = 3
\]
3. Divide both sides by \(3\):
\[
x = 1
\]
Now that we have \(x = 1\), we can substitute this value back into either of the original equations to find \(y\). We'll use \(y = 3x + 2\):
\[
y = 3(1) + 2 = 3 + 2 = 5
\]
Thus, the solution to the system of equations is the ordered pair \((1, 5)\).
Now, let's complete the table for the given \(x\) values:
| x | \(y = 6x - 1\) | \(y = 3x + 2\) |
|----|----------------|-----------------|
| -1 | \(y = 6(-1) - 1 = -6 - 1 = -7\) | \(y = 3(-1) + 2 = -3 + 2 = -1\) |
| 0 | \(y = 6(0) - 1 = 0 - 1 = -1\) | \(y = 3(0) + 2 = 0 + 2 = 2\) |
| 1 | \(y = 6(1) - 1 = 6 - 1 = 5\) | \(y = 3(1) + 2 = 3 + 2 = 5\) |
| 2 | \(y = 6(2) - 1 = 12 - 1 = 11\) | \(y = 3(2) + 2 = 6 + 2 = 8\) |
| 3 | \(y = 6(3) - 1 = 18 - 1 = 17\) | \(y = 3(3) + 2 = 9 + 2 = 11\) |
Now we can summarize the results in the table:
| x | \(y = 6x - 1\) | \(y = 3x + 2\) |
|----|----------------|-----------------|
| -1 | -7 | -1 |
| 0 | -1 | 2 |
| 1 | 5 | 5 |
| 2 | 11 | 8 |
| 3 | 17 | 11 |
The ordered pair that is a solution to the system of equations is \((1, 5)\).
\[
6x - 1 = 3x + 2
\]
Next, we solve for \(x\):
1. Subtract \(3x\) from both sides:
\[
6x - 3x - 1 = 2
\]
This simplifies to:
\[
3x - 1 = 2
\]
2. Add \(1\) to both sides:
\[
3x = 3
\]
3. Divide both sides by \(3\):
\[
x = 1
\]
Now that we have \(x = 1\), we can substitute this value back into either of the original equations to find \(y\). We'll use \(y = 3x + 2\):
\[
y = 3(1) + 2 = 3 + 2 = 5
\]
Thus, the solution to the system of equations is the ordered pair \((1, 5)\).
Now, let's complete the table for the given \(x\) values:
| x | \(y = 6x - 1\) | \(y = 3x + 2\) |
|----|----------------|-----------------|
| -1 | \(y = 6(-1) - 1 = -6 - 1 = -7\) | \(y = 3(-1) + 2 = -3 + 2 = -1\) |
| 0 | \(y = 6(0) - 1 = 0 - 1 = -1\) | \(y = 3(0) + 2 = 0 + 2 = 2\) |
| 1 | \(y = 6(1) - 1 = 6 - 1 = 5\) | \(y = 3(1) + 2 = 3 + 2 = 5\) |
| 2 | \(y = 6(2) - 1 = 12 - 1 = 11\) | \(y = 3(2) + 2 = 6 + 2 = 8\) |
| 3 | \(y = 6(3) - 1 = 18 - 1 = 17\) | \(y = 3(3) + 2 = 9 + 2 = 11\) |
Now we can summarize the results in the table:
| x | \(y = 6x - 1\) | \(y = 3x + 2\) |
|----|----------------|-----------------|
| -1 | -7 | -1 |
| 0 | -1 | 2 |
| 1 | 5 | 5 |
| 2 | 11 | 8 |
| 3 | 17 | 11 |
The ordered pair that is a solution to the system of equations is \((1, 5)\).
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