To solve the system of equations:
-
\(-3 + 6y = 6x\)
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\(2y - 14 = 4x\)
Let's start rearranging each equation to a more standard form.
From equation (1):
\[ 6y = 6x + 3 \] \[ y = x + \frac{1}{2} \quad \text{(divide by 6)} \]
From equation (2):
\[ 2y = 4x + 14 \] \[ y = 2x + 7 \quad \text{(divide by 2)} \]
Now we have a system of two expressions for \(y\):
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\(y = x + \frac{1}{2}\)
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\(y = 2x + 7\)
We can set the two expressions for \(y\) equal to each other:
\[ x + \frac{1}{2} = 2x + 7 \]
Now, solve for \(x\):
- Subtract \(x\) from both sides:
\[ \frac{1}{2} = x + 7 \]
- Subtract 7 from both sides:
\[ \frac{1}{2} - 7 = x \]
- Convert 7 to a fraction:
\[ \frac{1}{2} - \frac{14}{2} = x \]
This simplifies to:
\[ x = -\frac{13}{2} \]
Now that we have \(x\), we can find \(y\) by substituting \(x\) back into one of the equations. We'll use the first equation:
\[ y = x + \frac{1}{2} \]
Substituting in \(x = -\frac{13}{2}\):
\[ y = -\frac{13}{2} + \frac{1}{2} = -\frac{13}{2} + \frac{1}{2} = -\frac{12}{2} = -6 \]
Thus, the solution to the system of equations is:
\[ \boxed{\left(-\frac{13}{2}, -6\right)} \]