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 \(x\):
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Subtract \(3x\) from both sides: \[ 6x - 3x - 1 = 2 \] This simplifies to: \[ 3x - 1 = 2 \]
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Add \(1\) to both sides: \[ 3x = 3 \]
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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 \((1, 5)\).