How do you solve the system of equations xy = 12 and x^2 + y^2 = 40 using the substitution method?

Solving the System of Equations

To solve the system of equations given by:

From the first equation, we can express y in terms of x:

y = 12/x

Next, substitute this expression for y into the second equation:

x² + (12/x)² = 40

Now simplify:

x² + 144/x² = 40

Multiply through by x² to eliminate the fraction:

x⁴ - 40x² + 144 = 0

Let u = x². Then the equation becomes:

u² - 40u + 144 = 0

Now, we can use the quadratic formula u = (-b ± √(b² – 4ac)) / 2a:

u = (40 ± √((-40)² - 4 * 1 * 144)) / (2 * 1)

Calculating the discriminant:

u = (40 ± √(1600 - 576)) / 2

u = (40 ± √1024) / 2

u = (40 ± 32) / 2

u₁ = (72)/2 = 36

u₂ = (8)/2 = 4

Now substitute back to get the values for x:

x² = 36  ⇒  x = ±6

x² = 4   ⇒  x = ±2

Now, for each value of x, we can find y:

The solutions to the system of equations are: