To find the solution for the equations \(y = 2x - 1\) and \(y - 3 = 4x\), we will first rewrite the second equation in slope-intercept form (i.e., \(y = mx + b\)).
Starting with the equation \(y - 3 = 4x\):
\[ y = 4x + 3 \]
Now we have the two equations:
- \(y = 2x - 1\)
- \(y = 4x + 3\)
Next, we set them equal to each other to find the point of intersection:
\[ 2x - 1 = 4x + 3 \]
Now, solve for \(x\):
- Subtract \(2x\) from both sides:
\[ -1 = 2x + 3 \]
- Subtract 3 from both sides:
\[ -1 - 3 = 2x \] \[ -4 = 2x \]
- Divide by 2:
\[ x = -2 \]
Now, substitute \(x = -2\) back into one of the original equations to find \(y\). We can use the first equation \(y = 2x - 1\):
\[ y = 2(-2) - 1 \] \[ y = -4 - 1 \] \[ y = -5 \]
Therefore, the solution (point of intersection) of the two lines is:
\[ \boxed{(-2, -5)} \]
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