If the sumof two numbers is 4 and the sum of their squarea minus three times their product is 76, find the two numbers

Step-by-step explanation:

If the sumof two numbers is 4 and the sum of their squarea minus three times their product is 76, find the two numbers

Let us represent:.

First number = x

Second number = y

If the sum of two numbers is 4

= x + y = 4

x = 4 - y

The sum of their squares minus three times their product is 76

x² + y² - 3xy = 76

We substitute 4- y for x

(4 - y)² + y² - 3(4 - y)y = 76

(4 - y)(4 - y) + y² - (12 - 3y)y = 76

16 - 4y - 4y - y² + y² - 12y - 3y² = 76

16 -20y - 3y² = 76

3y² + 20y - 16 + 76 = 0

3y² + 20y + 60 = 0

the mistake in the pattern is that :

b. the decimal point was moved one too many place values each time

the correct pattern could be made by :

b. moving the decimal point to the right the same numbers of place values as the exponents

idk man good luck man