Let's solve the system of linear equations given by:
[tex]\[
\text{Equation 1:} \quad 3x + 2y = -4
\][/tex]
and
[tex]\[
\text{Equation 2:} \quad y = \frac{1}{2}x + 1
\][/tex]
We need to find the point that satisfies both equations.
### Verify the given points
#### Point 1: [tex]\((-1.5, 0.25)\)[/tex]
Substitute [tex]\((-1.5, 0.25)\)[/tex] into Equation 1:
[tex]\[ 3(-1.5) + 2(0.25) = -4 \][/tex]
[tex]\[ -4.5 + 0.5 = -4 \][/tex]
[tex]\[ -4 = -4 \quad \text{(True)} \][/tex]
Substitute [tex]\((-1.5, 0.25)\)[/tex] into Equation 2:
[tex]\[ 0.25 = \frac{1}{2}(-1.5) + 1 \][/tex]
[tex]\[ 0.25 = -0.75 + 1 \][/tex]
[tex]\[ 0.25 = 0.25 \quad \text{(True)} \][/tex]
Since [tex]\((-1.5, 0.25)\)[/tex] satisfies both equations, this point is a solution to the system of equations.
#### Point 2: [tex]\((-0.5, 0.75)\)[/tex]
Substitute [tex]\((-0.5, 0.75)\)[/tex] into Equation 1:
[tex]\[ 3(-0.5) + 2(0.75) = -4 \][/tex]
[tex]\[ -1.5 + 1.5 = -4 \][/tex]
[tex]\[ 0 = -4 \quad \text{(False)} \][/tex]
Since [tex]\((-0.5, 0.75)\)[/tex] does not satisfy Equation 1, it cannot be a solution to the system.
#### Point 3: [tex]\((1.5, 0.25)\)[/tex]
Substitute [tex]\((1.5, 0.25)\)[/tex] into Equation 1:
[tex]\[ 3(1.5) + 2(0.25) = -4 \][/tex]
[tex]\[ 4.5 + 0.5 = -4 \][/tex]
[tex]\[ 5 = -4 \quad \text{(False)} \][/tex]
Since [tex]\((1.5, 0.25)\)[/tex] does not satisfy Equation 1, it cannot be a solution to the system.
#### Point 4: [tex]\((-2.5, 1.5)\)[/tex]
Substitute [tex]\((-2.5, 1.5)\)[/tex] into Equation 1:
[tex]\[ 3(-2.5) + 2(1.5) = -4 \][/tex]
[tex]\[ -7.5 + 3 = -4 \][/tex]
[tex]\[ -4.5 = -4 \quad \text{(False)} \][/tex]
Since [tex]\((-2.5, 1.5)\)[/tex] does not satisfy Equation 1, it cannot be a solution to the system.
### Conclusion
The only point that satisfies both given equations is [tex]\((-1.5, 0.25)\)[/tex]. Thus, the solution to the system is [tex]\(\boxed{(-1.5, 0.25)}\)[/tex].
