AP DSC School Assistant Math Model Test 2

Concept: For a system of linear equations to be consistent (i.e., have at least one solution), the ratios of the coefficients of the corresponding variables in two equations must be equal. Here we compare the second and third equations because they involve only y and z. Explanation: Step 1: Write the given equations clearly. Equation (2): $(k+3)z = 3$ can be written as $0\cdot y + (k+3)z = 3$. Equation (3): $(2k+1)y + z = 2$. Step 2: For consistency, the ratio of the coefficients of y from equations (2) and (3) must equal the ratio of the coefficients of z from the same equations: $\displaystyle \frac{0}{2k+1} = \frac{k+3}{1}$ Step 3: Cross‑multiply: $0 \cdot 1 = (k+3)(2k+1)$ ⇒ $0 = 2k^2 + 6k + k + 3$ ⇒ $2k^2 + 7k + 3 = 0$ Step 4: Solve the quadratic using the formula $k = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ with $a=2$, $b=7$, $c=3$: Discriminant: $\Delta = 7^2 - 4(2)(3) = 49 - 24 = 25 > 0$ $k = \frac{-7 \pm \sqrt{25}}{2 \times 2} = \frac{-7 \pm 5}{4}$ Step 5: The two values are: $k = \frac{-7+5}{4} = \frac{-2}{4} = -\frac{1}{2}$ and $k = \frac{-7-5}{4} = \frac{-12}{4} = -3$. Answer: $k = -3$ and $k = -\frac{1}{2}$. Therefore, the correct option is A.