a) The linear equation is x - y = 15 and
the quadratic equation is x^2 - 20y = 300.
b) The numbers are x = 0, y = -15 and x = 20, y = 5
c) The two numbers that Simon is thinking of is x = 20, y = 5.
In this question,
Let the first number be x and
The second number be y
Part a:
The first number minus the second number is 15, then the linear equation is
⇒ x - y = 15 ------ (1)
The square of the first number minus 20 times the second number is equal to 300, then the quadratic equation is
⇒ x^2 - 20y = 300 ------- (2)
Part b:
Solve the system of equations
From (1) ⇒ y = x - 15 ------- (3)
Then, equation 2 becomes
⇒ x^2 - 20(x-15) = 300
⇒ x^2 - 20x + 300 = 300
⇒ x^2 - 20x = 300 - 300
⇒ x^2 - 20x = 0
⇒ x(x-20) = 0
⇒ x = 0 or x = 20
When x = 0, y = -15 and
When x = 20, y = 5
Part c:
The two numbers that Simon is thinking of is
The value of x and y are positive.
So, x = 20 and y = 5.
Hence we can conclude that
a) The linear equation is x - y = 15 and
the quadratic equation is x^2 - 20y = 300.
b) The values are x = 0, y = -15 and x = 20, y = 5
c) The two numbers that Simon is thinking of is x = 20, y = 5.
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