Question

!! your goal is to earn 10 points. 3. a) how many points of intersection does the given system have? \\(\

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

\\) b) what is the solution (x, y, z) of the following system: \\(\

$$\begin{cases} 4x - 2y + 3z = 9 \\\\ x - 2y = -3 \\\\ 2x + 3y = 1 \\end{cases}$$

\\) 2 points (a.rei.c.6/a.rei.c.7) 6. for what value of m is the equation true? \\(x^2 + mx - 36 = (x - 3)(x + 12)\\) 1 point (a.sse.a.2)

Explanation:

3a)

Step1: Set equations equal

To find intersection points, set \(1 - x^2 = x - 1\).

Step2: Rearrange into quadratic

Rearrange to \(x^2 + x - 2 = 0\).

Step3: Solve quadratic

Factor: \((x + 2)(x - 1)=0\), so \(x=-2\) or \(x = 1\). Two solutions, so two intersection points.

Step1: Solve \(x - 2y=-3\) and \(2x + 3y = 1\)

From \(x - 2y=-3\), get \(x = 2y - 3\). Substitute into \(2x + 3y = 1\): \(2(2y - 3)+3y = 1\) → \(4y - 6 + 3y = 1\) → \(7y = 7\) → \(y = 1\). Then \(x = 2(1)-3=-1\).

Step2: Substitute \(x=-1,y = 1\) into \(4x - 2y + 3z = 9\)

\(4(-1)-2(1)+3z = 9\) → \(-4 - 2 + 3z = 9\) → \(3z = 15\) → \(z = 5\).

Step1: Expand \((x - 3)(x + 12)\)

Expand: \(x^2 + 12x - 3x - 36=x^2 + 9x - 36\).

Step2: Compare with \(x^2 + mx - 36\)

So \(m = 9\).

Answer:

2

3b)