Distribute x in the equation x(4 - x) = 16 to get 4x - x² = 16. Rearrange this into the standard quadratic form ax² + bx + c = 0 by moving all terms to one side, resulting in x² - 4x + 16 = 0.

To find the solutions for x, we can use the quadratic formula. First, we calculate the discriminant, Delta = b² - 4ac. For the equation x² - 4x + 16 = 0, we have a=1, b=-4, and c=16. Substituting these values, we get Delta = (-4)² - 4(1)(16) = 16 - 64 = -48.

To find the solutions for x, we can use the quadratic formula. First, we calculate the discriminant, Delta = b² - 4ac. For the equation x² - 4x + 16 = 0, we have a=1, b=-4, and c=16. Substituting these values, we get Delta = (-4)² - 4(1)(16) = 16 - 64 = -48.

Since the discriminant Delta = -48 is less than zero, the quadratic equation x² - 4x + 16 = 0 has no real roots. This means there are no real values of x and y that satisfy both given equations simultaneously.

Since the discriminant Delta = -48 is less than zero, the quadratic equation x² - 4x + 16 = 0 has no real roots. This means there are no real values of x and y that satisfy both given equations simultaneously.

$x + y = 4$ $xy = 16$ (system of equations, variables, algebra)

Answer: There are no real solutions for $x$ and $y$ .

Step-by-step solution

Step 1: Express one variable in terms of the other

We are given two equations: $x + y = 4$ and $xy = 16$ . To solve this system, we can use the substitution method. From the first equation, $x + y = 4$ , we can express $y$ in terms of $x$ by subtracting $x$ from both sides.

Step 2: Substitute into the second equation

Now, substitute the expression for $y$ from the first step into the second equation, $xy = 16$ . This will give us an equation with only one variable, $x$ .

Step 3: Form a quadratic equation

Distribute $x$ in the equation $x(4 - x) = 16$ to get $4x - x^2 = 16$ . Rearrange this into the standard quadratic form $ax^2 + bx + c = 0$ by moving all terms to one side, resulting in $x^2 - 4x + 16 = 0$ .

Step 4: Calculate the discriminant

To find the solutions for $x$ , we can use the quadratic formula. First, we calculate the discriminant, $\Delta = b^2 - 4ac$ . For the equation $x^2 - 4x + 16 = 0$ , we have $a=1$ , $b=-4$ , and $c=16$ . Substituting these values, we get $\Delta = (-4)^2 - 4(1)(16) = 16 - 64 = -48$ .

Step 5: Determine the nature of the roots

Since the discriminant $\Delta = -48$ is less than zero, the quadratic equation $x^2 - 4x + 16 = 0$ has no real roots. This means there are no real values of $x$ and $y$ that satisfy both given equations simultaneously.

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x+y=4x + y = 4x+y=4 xy=16xy = 16xy=16 (system of equations, variables, algebra)