Linear Equations, Inequalities and Systems: math test solutions, Grade 11 – download the PDF

1 Lines and slope / 3 pts

(a) \(m = \dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6} = -1\). (1 pt)

(b) \(y - 5 = -(x + 2)\), so \(y = -x + 3\). Check with \(B\): \(-4 + 3 = -1\). (1 pt)

(c) Same slope \(-1\) and y-intercept \(-4\): \(y = -x - 4\). (1 pt)

2 Absolute value / 3 pts

(a) \(2x - 5 = 11\) gives \(x = 8\); \(2x - 5 = -11\) gives \(x = -3\). Solutions: \(8\) and \(-3\). (1 pt)

(b) \(x + 1 \ge 5\) or \(x + 1 \le -5\), so \(x \ge 4\) or \(x \le -6\). In interval notation, \((-\infty, -6] \cup [4, \infty)\). (2 pts)

3 Compound inequalities / 3 pts

(a) Subtract \(3\): \(-12 \le 2x < 4\). Divide by \(2\): \(-6 \le x < 2\), that is \([-6, 2)\). (2 pts)

(b) \(4x < 4\) gives \(x < 1\); \(2x \ge 10\) gives \(x \ge 5\). The solution is \((-\infty, 1) \cup [5, \infty)\). (1 pt)

4 Concert tickets / 3 pts

(a) \(s + a = 220\) and \(9s + 14a = 2380\). (1 pt)

(b) Substitute \(s = 220 - a\): \(1980 - 9a + 14a = 2380\), so \(5a = 400\) and \(a = 80\), \(s = 140\). (1 pt) Check: \(9(140) + 14(80) = 1260 + 1120 = 2380\). (0.5 pt) The concert sold 140 student tickets and 80 adult tickets. (0.5 pt)

5 A system of three equations / 3 pts

Add the first two equations: \(2x + 3z = 18\). (1 pt)

Subtract the third equation from the first: \(-x + 2z = 5\), so \(x = 2z - 5\). Substitute: \(4z - 10 + 3z = 18\), so \(z = 4\) and \(x = 3\). (1 pt)

From the first equation, \(y = 9 - 3 - 4 = 2\). Check in the second: \(3 - 2 + 8 = 9\); in the third: \(6 + 2 - 4 = 4\). The solution is \((3, 2, 4)\). (1 pt)

6 Linear programming / 3 pts

The vertices are \((0, 0)\), \((10, 0)\), \((0, 6)\), and the intersection of the two lines: subtracting the equations gives \(2y = 8\), so \(y = 4\) and \(x = 6\): \((6, 4)\). (2 pts)

\(P(0, 0) = 0\), \(P(10, 0) = 40\), \(P(0, 6) = 30\), \(P(6, 4) = 24 + 20 = 44\). The maximum is \(44\), reached at \((6, 4)\). (1 pt)

7 Matrices / 2 pts

(a) \(\det A = 4 \cdot 1 - 3 \cdot 1 = 1\), so \(A^{-1} = \begin{bmatrix} 1 & -3 \\ -1 & 4 \end{bmatrix}\). (1 pt)

(b) \(X = A^{-1}B = \begin{bmatrix} 17 - 15 \\ -17 + 20 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}\). Check: \(8 + 9 = 17\) and \(2 + 3 = 5\). So \(x = 2\), \(y = 3\). (1 pt)