To find the solution to the system of equations \(y = 2x - 1\) and \(y - 3 = 4x\), we first need to rewrite the second equation in the slope-intercept form \(y = mx + b\).
Starting with the second equation:
\[ y - 3 = 4x \]
Adding 3 to both sides gives:
\[ y = 4x + 3 \]
Now we have the two equations to work with:
- \(y = 2x - 1\)
- \(y = 4x + 3\)
Next, we can set the two equations equal to each other to find the intersection point:
\[ 2x - 1 = 4x + 3 \]
Now, we'll solve for \(x\). First, subtract \(2x\) from both sides:
\[ -1 = 2x + 3 \]
Next, subtract 3 from both sides:
\[ -1 - 3 = 2x \]
This simplifies to:
\[ -4 = 2x \]
Now divide both sides by 2:
\[ x = -2 \]
Now that we have \(x\), we can substitute it back into either equation to find \(y\). We'll use the first equation:
\[ y = 2(-2) - 1 \] \[ y = -4 - 1 \] \[ y = -5 \]
So the solution to the system of equations is \((-2, -5)\).
Thus, the ordered pair \((-2, -5)\) is the solution to the system of equations because it is the point at which both lines intersect, so both lines share this point.
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