Respuesta :
Answer:
Step-by-step explanation:
Part A, the system of equation is, presented here as follows;
(1) 2·y = 3·x + 2
(2) y = 2·x² - 3·x + 2
Part B; The graph of the system of equation is attached
The reason for the above answers is as follows;
Part A
A system of equations are a given number of equations from which a common solution of the system can be found
A linear equation is an algebraic equation in which the maximum exponent of the variables is one and the graph of the equation is a straight line which is of the form y = m·x + c
A maximum value of the exponents of the variable quadratic equation is 2, and the general form of the quadratic equation is y = a·x² + b·x + c, where a, b, and c are real numbers
Therefore, the system of equation that can be created is as follows;
(1) 2·y = 3·x + 2
(2) y = 2·x² - 3·x + 2
Method for solving;
Divide equation (1) by 2, and equate both values of y to find the common solution as follows;
2·y/2 = (3·x + 2)/2 = 1.5·x + 1
∴ y = 1.5·x + 1
Equating both values of y gives;
y = 1.5·x + 1
y = 2·x² - 3·x + 2
Therefore;
1.5·x + 1 = 2·x² - 3·x + 2
2·x² - 3·x - 1.5·x + 2 - 1 = 0
2·x² - 4.5·x + 1 = 0
Using the quadratic formula, we get;
x = (4.5 ± √((-4.5)² - 4×2×1))/(2 × 2)
x = 2, or x = 0.25
From which we get;
y = 1.5 × 2 + 1 = 4, or y = 1.5 × 0.25 + 1 = 1.375
The points where the line graph and the quadratic graph intersect are;
(2, 4), and (0.25, 1.375)
Part 2
