Answer:
a. Yes, (-3 , -5) is a solution to the system because the graphs of the two equations intersected at it
b. No, (3 , 1) is not a solution to the system because the graphs of the equations did not intersect at it, they intersected at point (1 , 3)
Step-by-step explanation:
The graph has a line which represents an linear equation y = mx + b, where m is the slope of the line and b is the y-intercept (value y at x = 0)
∵ The line intersects the y-axis at point (0 , 1)
∴ b = 1
∵ The line passes through points (0 , 1) and (1 , 3)
∵ m = Δy/Δx
∴ m = [tex]\frac{3-1}{1-0}[/tex]
∴ m = 2
The equation of the line is y = 2 x + 1 ⇒ (1)
The graph has a downward parabola which represents a quadratic equation y = ax² + bx + c, where
- a = -1 because the parabola is down ward
- b = 0 because the vertex of the parabola is on the y-axis which means x = 0
- c = 4 because c is the y-intercept
The quadratic equation is y = -x² + 4 ⇒ (2)
The solutions of the system of equations are the intersection points between the line and the parabola
∵ The points of intersections are (1 , 3) and (-3 , -5)
∴ The solutions of the system of equations are (1 , 3) and (-3 , -5)
Let us verify them algebraically
Equate equations (1) and (2)
∵ 2x + 1 = -x² + 4
- Add x² to both sides
∴ x² + 2x + 1 = 4
- Subtract 4 from both sides
∴ x² + 2x - 3 = 0
- Factorize it into two factors
∴ (x + 3)(x - 1) = 0
- Equate each factor by 0
∵ x + 3 = 0
- subtract 3 from both sides
∴ x = -3
OR
∵ x - 1 = 0
- Add 1 to both sides
∴ x = 1
Substitute the values of x in equation (2) to find the values of y
∵ y = 2(-3) + 1
∴ y = -6 + 1
∴ y = -5
OR
∵ y = 2(1) + 1
∴ y = 2 + 1
∴ y = 3
∴ The solutions are (-3 , -5) and (1 , 3)
Lets answer the questions in a and b
a. Yes, (-3 , -5) is a solution to the system because the graphs of the two equations intersected at it
b. No, (3 , 1) is not a solution to the system because the graphs of the equations did not intersect at it, they intersected at point (1 , 3)
