Question

How many real solutions exist for this system of equations? y = x^2 + 4 y = 4x

Answer 1

Answer:

This system of equations has one real solution, (2, 8).

Step-by-step explanation:

Given:

y = x² + 4

y = 4x

1. Substitute the given value of y into y = x² + 4:

⇒ 4x = x² + 4

2. Solve for x (by factoring):

⇒ 4x = x² + 4

⇒ 0 = x² - 4x + 4

⇒ 0 = (x - 2)² [perfect square rule: a² -2ab + b² = (a - b)²]

⇒ 0 = (x - 2)² [take the square root of both sides]

⇒ $\sqrt{0} = \sqrt{(x - 2)^2}$

⇒ 0 = x - 2 [add 2 to both sides]

⇒ 0 + 2 = x - 2 + 2

⇒ 2 = x

3. Find the value of y by substituting the given value of x into y = 4x:

⇒ y = 4(2)

⇒ y = 8

4. Check your work:

⇒ y = x² + 4          and          ⇒ y = 4x

⇒ 8 = 2² + 4         and          ⇒ 8 = 4(2)

⇒ 8 = 4 + 4          and           ⇒ 8 = 8 ✔

⇒ 8 = 8 ✔

Solution: (x, y) ⇒ (2, 8)

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