Answer:
3, 12
Step-by-step explanation:
Et x and y be the required integers.
Case 1: x = 5y - 3...(1)
Case 2: xy = 36
Hence, (5y - 3)*y = 36
[tex]5 {y}^{2} - 3y = 36 \\ 5 {y}^{2} - 3y - 36 = 0 \\ 5 {y}^{2} - 15y + 12y - 36 = 0 \\ 5y(y - 3) + 12(y - 3) = 0 \\ (y - 3)(5y + 12) = 0 \\ y - 3 = 0 \: or \: 5y + 12 = 0 \\ y = 3 \: \: or \: \: y = - \frac{12}{5} \\ \because \: y \in \: I \implies \: y \neq - \frac{12}{5} \\ \huge \purple{ \boxed{ \therefore \: y = 3}} \\ \because \: x = 5y - 3..(equation \: 1) \\ \therefore \: x = 5 \times 3 - 3 = 15 - 3 = 12 \\ \huge \red{ \boxed{ x = 12}}[/tex]
Hence, the required integers are 3 and 12.
let
x = one integer
y = another integer
x = 5y - 3
If the product of the two integers is 36, then find the integers.
x * y = 36
(5y - 3) * y = 36
5y² - 3y = 36
5y² - 3y - 36 = 0
Solve the quadratic equation using factorization method
That is, find two numbers whose product will give -180 and sum will give -3
Note: coefficient of y² multiplied by -36 = -180
5y² - 3y - 36 = 0
The numbers are -15 and +12
5y² - 15y + 12y - 36 = 0
5y(y - 3) + 12 (y - 3) = 0
(5y + 12) (y - 3) = 0
5y + 12 = 0 y - 3 = 0
5y = - 12 y = 3
y = -12/5
The value of y can not be negative
Therefore,
y = 3
Substitute y = 3 into x = 5y - 3
x = 5y - 3
x = 5(3) - 3
= 15 - 3
= 12
x = 12
Therefore,
(x, y) = (12, 3)
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