To determine the number of solutions for the given system of linear equations, we need to analyze the equations:
- \(-x + y = 3\) (Equation 1)
- \(-2x + 2y = 3\) (Equation 2)
We will solve the system by rewriting and manipulating the equations step by step.
First, let's rewrite the first equation in slope-intercept form (y = mx + b):
From Equation 1: \[ y = x + 3 \]
Next, we simplify Equation 2. We can divide the entire equation by 2:
\[ -2x + 2y = 3 \implies -x + y = \frac{3}{2} \]
Now we can write both equations in a comparable form:
- \(y = x + 3\) (from Equation 1)
- \(y = x + \frac{3}{2}\) (from the simplified Equation 2)
Next, we examine the slopes and the y-intercepts of these two equations:
- The first equation, \(y = x + 3\), has a slope of \(1\) and a y-intercept of \(3\).
- The second equation, \(y = x + \frac{3}{2}\), also has a slope of \(1\) but has a different y-intercept of \(\frac{3}{2}\).
Since both lines have the same slope but different y-intercepts, they are parallel lines. Parallel lines do not intersect at any point; therefore, there are no solutions to the system of equations.
Thus, the conclusion is: \[ \text{No Solutions} \]