Question
Ms. Mac asked her students to determine a solution for 2z5y = 9
Bella believes (-2, 1) is a solution to the equation.
Marco believes (11, 2) is a solution to the equation.
Liam believes (2,-1)is a solution to the equation.
(1 point)
named a point that is a solution that lies on the graph of the equation
named a point that is not a solution and lies above the graph of the equation.
inamed a
point that is not a solution and lies below the graph of the equation.
#Marco
Liam
#Bella
Bella believes (-2, 1) is a solution to the equation.
Marco believes (11, 2) is a solution to the equation.
Liam believes (2,-1)is a solution to the equation.
(1 point)
named a point that is a solution that lies on the graph of the equation
named a point that is not a solution and lies above the graph of the equation.
inamed a
point that is not a solution and lies below the graph of the equation.
#Marco
Liam
#Bella
Answers
GPT-4o mini
To determine which students' proposed solutions are correct, substitute the provided points into the equation \( 2z + 5y = 9 \).
1. **Testing Bella's point (-2, 1)**:
\[
2(-2) + 5(1) = -4 + 5 = 1 \quad \text{(not a solution)}
\]
2. **Testing Marco's point (11, 2)**:
\[
2(11) + 5(2) = 22 + 10 = 32 \quad \text{(not a solution)}
\]
3. **Testing Liam's point (2, -1)**:
\[
2(2) + 5(-1) = 4 - 5 = -1 \quad \text{(not a solution)}
\]
Since none of the proposed solutions are valid, we need to find points that lie on the line \( 2z + 5y = 9 \).
### Named a point that is a solution:
To find a valid solution, we can set \( z = 0 \) and solve for \( y \):
\[
2(0) + 5y = 9 \implies 5y = 9 \implies y = \frac{9}{5} \quad \text{(valid point: (0, 1.8))}
\]
### Named a point that is not a solution and lies above the graph:
Any point above the line \( 2z + 5y = 9 \) will yield a value greater than 9 when substituted. For example, let's use the point (0, 3):
\[
2(0) + 5(3) = 0 + 15 = 15 \quad \text{(not a solution and lies above)}
\]
### Named a point that is not a solution and lies below the graph:
Any point below the line \( 2z + 5y = 9 \) will yield a value less than 9. For example, let's use the point (0, 1):
\[
2(0) + 5(1) = 0 + 5 = 5 \quad \text{(not a solution and lies below)}
\]
To summarize:
1. A point that is a solution: (0, 1.8)
2. A point that is not a solution and lies above: (0, 3)
3. A point that is not a solution and lies below: (0, 1)
1. **Testing Bella's point (-2, 1)**:
\[
2(-2) + 5(1) = -4 + 5 = 1 \quad \text{(not a solution)}
\]
2. **Testing Marco's point (11, 2)**:
\[
2(11) + 5(2) = 22 + 10 = 32 \quad \text{(not a solution)}
\]
3. **Testing Liam's point (2, -1)**:
\[
2(2) + 5(-1) = 4 - 5 = -1 \quad \text{(not a solution)}
\]
Since none of the proposed solutions are valid, we need to find points that lie on the line \( 2z + 5y = 9 \).
### Named a point that is a solution:
To find a valid solution, we can set \( z = 0 \) and solve for \( y \):
\[
2(0) + 5y = 9 \implies 5y = 9 \implies y = \frac{9}{5} \quad \text{(valid point: (0, 1.8))}
\]
### Named a point that is not a solution and lies above the graph:
Any point above the line \( 2z + 5y = 9 \) will yield a value greater than 9 when substituted. For example, let's use the point (0, 3):
\[
2(0) + 5(3) = 0 + 15 = 15 \quad \text{(not a solution and lies above)}
\]
### Named a point that is not a solution and lies below the graph:
Any point below the line \( 2z + 5y = 9 \) will yield a value less than 9. For example, let's use the point (0, 1):
\[
2(0) + 5(1) = 0 + 5 = 5 \quad \text{(not a solution and lies below)}
\]
To summarize:
1. A point that is a solution: (0, 1.8)
2. A point that is not a solution and lies above: (0, 3)
3. A point that is not a solution and lies below: (0, 1)