Question

find the solution of each system of equations. 9. \\(\

$$\begin{cases} y = x^2 + 3x + 1 \\\\ y = -x + 1 \\end{cases}$$

\\) 10. \\(\

$$\begin{cases} y = x^2 + 1 \\\\ y = -2x \\end{cases}$$

\\)

Explanation:

Problem 9

Step1: Set equations equal

Since \( y = x^2 + 3x + 1 \) and \( y = -x + 1 \), set \( x^2 + 3x + 1 = -x + 1 \).

Step2: Simplify equation

Subtract \( -x + 1 \) from both sides: \( x^2 + 4x = 0 \). Factor: \( x(x + 4) = 0 \).

Step3: Solve for x

\( x = 0 \) or \( x + 4 = 0 \Rightarrow x = -4 \).

Step4: Find corresponding y

For \( x = 0 \), \( y = -0 + 1 = 1 \). For \( x = -4 \), \( y = -(-4) + 1 = 5 \).

Step1: Set equations equal

Since \( y = x^2 + 1 \) and \( y = -2x \), set \( x^2 + 1 = -2x \).

Step2: Simplify equation

Rearrange: \( x^2 + 2x + 1 = 0 \). Factor: \( (x + 1)^2 = 0 \).

Step3: Solve for x

\( x + 1 = 0 \Rightarrow x = -1 \).

Step4: Find corresponding y

For \( x = -1 \), \( y = -2(-1) = 2 \).

Answer:

Solutions are \( (0, 1) \) and \( (-4, 5) \)

Problem 10