Answer :
Sure! Let's determine which of these given points satisfy the given linear equation [tex]\(y = -\frac{1}{2}x + 1\)[/tex]. We will substitute each [tex]\((x, y)\)[/tex] pair into the equation to see if the equality holds true.
1. For the point [tex]\((-2, 2)\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-2) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 1 + 1 = 2 \][/tex]
Since the left-hand side [tex]\(y = 2\)[/tex] matches the right-hand side, the point [tex]\((-2, 2)\)[/tex] is a solution to the equation.
2. For the point [tex]\((4, -1)\)[/tex]:
Substitute [tex]\(x = 4\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(4) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = -2 + 1 = -1 \][/tex]
Since the left-hand side [tex]\(y = -1\)[/tex] matches the right-hand side, the point [tex]\((4, -1)\)[/tex] is a solution to the equation.
3. For the point [tex]\((0, 4)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(0) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 0 + 1 = 1 \][/tex]
Since the left-hand side [tex]\(y = 4\)[/tex] does not match the right-hand side, the point [tex]\((0, 4)\)[/tex] is not a solution to the equation.
4. For the point [tex]\((0, 1)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(0) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 0 + 1 = 1 \][/tex]
Since the left-hand side [tex]\(y = 1\)[/tex] matches the right-hand side, the point [tex]\((0, 1)\)[/tex] is a solution to the equation.
5. For the point [tex]\((-1, 1)\)[/tex]:
Substitute [tex]\(x = -1\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-1) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = \frac{1}{2} + 1 = \frac{3}{2} \][/tex]
Since the left-hand side [tex]\(y = 1\)[/tex] does not match the right-hand side, the point [tex]\((-1, 1)\)[/tex] is not a solution to the equation.
6. For the point [tex]\((-2, -2)\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-2) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 1 + 1 = 2 \][/tex]
Since the left-hand side [tex]\(y = -2\)[/tex] does not match the right-hand side, the point [tex]\((-2, -2)\)[/tex] is not a solution to the equation.
After checking all the points, the points that satisfy the equation [tex]\(y = -\frac{1}{2}x + 1\)[/tex] are:
[tex]\[ (-2, 2), (4, -1), (0, 1) \][/tex]
1. For the point [tex]\((-2, 2)\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-2) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 1 + 1 = 2 \][/tex]
Since the left-hand side [tex]\(y = 2\)[/tex] matches the right-hand side, the point [tex]\((-2, 2)\)[/tex] is a solution to the equation.
2. For the point [tex]\((4, -1)\)[/tex]:
Substitute [tex]\(x = 4\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(4) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = -2 + 1 = -1 \][/tex]
Since the left-hand side [tex]\(y = -1\)[/tex] matches the right-hand side, the point [tex]\((4, -1)\)[/tex] is a solution to the equation.
3. For the point [tex]\((0, 4)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(0) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 0 + 1 = 1 \][/tex]
Since the left-hand side [tex]\(y = 4\)[/tex] does not match the right-hand side, the point [tex]\((0, 4)\)[/tex] is not a solution to the equation.
4. For the point [tex]\((0, 1)\)[/tex]:
Substitute [tex]\(x = 0\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(0) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 0 + 1 = 1 \][/tex]
Since the left-hand side [tex]\(y = 1\)[/tex] matches the right-hand side, the point [tex]\((0, 1)\)[/tex] is a solution to the equation.
5. For the point [tex]\((-1, 1)\)[/tex]:
Substitute [tex]\(x = -1\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-1) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = \frac{1}{2} + 1 = \frac{3}{2} \][/tex]
Since the left-hand side [tex]\(y = 1\)[/tex] does not match the right-hand side, the point [tex]\((-1, 1)\)[/tex] is not a solution to the equation.
6. For the point [tex]\((-2, -2)\)[/tex]:
Substitute [tex]\(x = -2\)[/tex] into the equation:
[tex]\[ y = -\frac{1}{2}(-2) + 1 \][/tex]
Simplifying this, we get:
[tex]\[ y = 1 + 1 = 2 \][/tex]
Since the left-hand side [tex]\(y = -2\)[/tex] does not match the right-hand side, the point [tex]\((-2, -2)\)[/tex] is not a solution to the equation.
After checking all the points, the points that satisfy the equation [tex]\(y = -\frac{1}{2}x + 1\)[/tex] are:
[tex]\[ (-2, 2), (4, -1), (0, 1) \][/tex]
