Question:

<p><strong>Step 1:</strong> Form the coefficient matrix and find when det(A) = 0.</p><p>$$A = \begin{pmatrix} k+2 & 10 \\ k & k+3 \end{pmatrix}$$</p><p>$$\det(A) = (k+2)(k+3) - 10k = k^2 + 5k + 6 - 10k = k^2 - 5k + 6$$</p><p>$$= (k-2)(k-3)$$</p><p>So det(A) = 0 when <strong>k = 2 or k = 3</strong></p><p><strong>Step 2:</strong> Check consistency for each value using Cramer's rule conditions.</p><p><strong>For k = 2:</strong> System becomes:</p><p>$$4x + 10y = 2 \implies 2x + 5y = 1$$</p><p>$$2x + 5y = 1$$</p><p>Both equations are identical → infinitely many solutions ✗</p><p><strong>For k = 3:</strong> System becomes:</p><p>$$5x + 10y = 3$$</p><p>$$3x + 6y = 2$$</p><p>From equation 1: multiply by 3/5 gives $3x + 6y = 9/5$</p><p>From equation 2: $3x + 6y = 2$</p><p>Since $9/5 ≠ 2$, the system is <strong>inconsistent</strong> → no solution ✓</p><p><strong>Step 3:</strong> Verify using the condition: When det(A) = 0, system has no solution if $\frac{Δ_x}{Δ}$ or $\frac{Δ_y}{Δ}$ is undefined (0/0 avoided but equations conflict).</p><p>∴ Answer: <strong>1</strong> (only k = 3 gives no solution)</p>