The difference between 2 positive numbers, x and y where x > y, is 7 and the sum of their squares is 137. By forming 2 equations in x and y, find the product of these 2 numbers.
We are given
x²+y²=137
(x-y)=7
this means
49=(x-y)²
=x²-2xy+y²
=x²+y²-2xy
=137-2xy
From
49=137-2xy
can you solve for xy?
The longer way:
equation 1:
x-y=7
equation 2:
x²+y²=137
Now solve the non-linear system in x and y.
To find the product of the two numbers, we need to solve the system of equations:
Equation 1: x - y = 7
Equation 2: x^2 + y^2 = 137
We can solve this system of equations algebraically by substitution or elimination. Let's use the substitution method.
From Equation 1, we can rearrange it to get x = y + 7. Now substitute this expression for x in Equation 2:
(y + 7)^2 + y^2 = 137
Expanding the square and simplifying the expression, we get:
y^2 + 14y + 49 + y^2 = 137
2y^2 + 14y + 49 = 137
2y^2 + 14y - 88 = 0
Divide both sides of the equation by 2 to simplify it further:
y^2 + 7y - 44 = 0
Now we can factor this quadratic equation:
(y - 4)(y + 11) = 0
Setting each factor equal to zero and solving for y, we get two possible values for y:
y - 4 = 0 --> y = 4
y + 11 = 0 --> y = -11
Since we are looking for positive numbers, we discard the negative value of y.
Now substitute the value of y = 4 back into Equation 1 to find x:
x - 4 = 7 --> x = 11
So the two positive numbers are x = 11 and y = 4.
To find their product, multiply these numbers together:
Product = 11 * 4 = 44
Therefore, the product of the two numbers is 44.