Solve the system by the substitution method: xy = 12, x² + y² = 40.

Solution:

We have to use the method of substitution to solve the problem. 

We have equations:

⇒ xy = 12              ------(1)

⇒ x² + y² = 40       ------(2)

Hence, we write equation (1) as y in terms of x, i.e, 

⇒ y = 12/x             ------(3)

Now, we substitute the value of y from equation (3) to equation (2).

⇒ x² + 144/x² = 40 

⇒ x⁴ - 40x² + 144 = 0

Now, we solve for x in the above quadratic equation. Now let's take z = x²   ------(4)

⇒ z² - 40z + 144 = 0

We use the quadratic formula to solve the equation above. After solving the equation, we get z = 4 and z = 36.

Hence, we get x = -2, 2 from z = 4, and x = 6, -6 from z = 36.

Now, after we substitute these values in equation (1), we get the following values as our solutions:

⇒ x = 2, y = 6

⇒ x = -2, y = -6

⇒ x = 6, y = 2

⇒ x = -6, y = -2

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Solve the system by the substitution method. xy = 12 and x² + y² = 40

Summary:

The system by the substitution method. xy = 12 and x² + y² = 40 is x = ±6 and y = ±2