Question

lesson 3:rsg
20.
\

$$\begin{cases} y = 2x + 12 \\ y = 10x - x^2 \\end{cases}$$

Explanation:

Step1: Set the two equations equal

Since both equations are solved for \( y \), we can set them equal to each other: \( 2x + 12 = 10x - x^2 \)

Step2: Rearrange into standard quadratic form

Bring all terms to one side: \( x^2 - 8x + 12 = 0 \)

Step3: Factor the quadratic equation

Factor: \( (x - 2)(x - 6) = 0 \)

Step4: Solve for \( x \)

Set each factor equal to zero: \( x - 2 = 0 \) or \( x - 6 = 0 \), so \( x = 2 \) or \( x = 6 \)

Step5: Find corresponding \( y \) values

For \( x = 2 \): \( y = 2(2) + 12 = 16 \)
For \( x = 6 \): \( y = 2(6) + 12 = 24 \)

Answer:

The solutions are \( (2, 16) \) and \( (6, 24) \)